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Aug 9, 2008, 12:04:00 AM8/9/08

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The multiplication rules for the seven non-scalar Octonion

basis units are best described by seven permutations, any

of which when combined with the scalar basis unit form a

quaternion subalgebra.

basis units are best described by seven permutations, any

of which when combined with the scalar basis unit form a

quaternion subalgebra.

I would like to use the short-hand notation (ijk) to imply

the permutation basis unit multiplication rules for i,j,k

not equal to each other and not equal to 0

ui * uj = uk uj * uk = ui uk * ui = uj

uj * ui = -uk uk * uj = -ui ui * uk = -uj

(ijk) == (jki) == (kij) equivalent cyclic permutations

Each of the seven permutations define six multiplication

table entries, so 42 of 64 positions of the Octonion

multiplication table are covered by permutation rules.

The variability in the set of possible Octonion algebras is

entirely covered by the juxtaposition of units within the

seven permutations, since for any algebra the remaining 22

basis unit products are singularly defined for all Octonion

algebras for i not 0 as

ui * ui = -u0

u0 * ui = ui * u0 = ui

u0 * u0 = u0

Each ui (i not 0) appears in three of the seven

permutations. The nature of a permutation triplet is such

that the permutation multiplication rules above are negated

by the exchange in position of any two members. Any second

exchange restores the permutation to its original

configuration.

The three negations of (ijk) are

(jik) exchange ij

(kji) exchange ik

(ikj) exchange jk

The equivalent cyclic permutations for the negation are

(jik) == (kji) == (ikj)

If we do not allow unit aliasing, we must insist all

attempts to modify Octonion Algebra do not alter the

triplet of units in the set of permutations. That is to say

if we started with (123) we do not end up with (124). We

will only allow the negation of one or more permutations as

a possible change to the multiplication rules. This will

set the maximum number of different algebra definitions to

128. However, not all 128 of these moves result in valid

Octonion Algebra.

For "/a" representing the conjugate of Octonion "a" and for

"b" also Octonion we must have

/a * ( a * b ) = (/a * a) * b

If we check this equality against the 128 possible rules,

we find only sixteen work, and therefore are valid Octonion

rules. As will be shown, the sixteen fall into two sets of

eight I call "Left Octonion Algebra" and "Right Octonion

Algebra". The members of each type are all algebraically

isomorphic, but Left is not algebraically isomorphic to

Right.

Isomorphism is an overloaded term in mathematics. Perhaps

it would be helpful to discuss what is meant by the phrase

"algebraically isomorphic". Two algebras are isomorphic if

and only if their multiplication tables are equivalent.

"Equivalent" does not mean identical. A row pair may be

exchanged followed by the exchange of same column pair,

changing the appearance of the table without changing the

product definition of any two basis units, the two tables

are "equivalent".

The distinction between "Left" and "Right" Octonion Algebra

is clearly seen within the three permutations that include

any one of the seven basis units. Cyclically shift these

permutations such that the common basis unit is in the

central position. The three basis units on the right end of

these three permutations will be members of a fourth

permutation and those on the left end will not for "Right"

Octonion Algebra. Similarly, the three basis units on the

left end will be members of a fourth permutation and those

on the right end will not for "Left" Octonion Algebra.

The partitioning between the three permutations that

include one of the basis units and the four that do not is

important for understanding the permutation negations that

map within or between Left and Right Octonion Algebra

types. To see this, let us examine the seven permutations

assuming one is (ijk).

Within the three permutations including basis unit ui,

basis units uj and uk are only present in permutation

(ijk). Now consider the four permutations that do not

contain ui. Two of four will include the basis unit uj and

not uk, the other two will include the basis unit uk and

not uj.

For instance, take (ijk) = (123) in the following Right

Octonion Algebra

(312) has 1,2,3

(415) has 1

(617) has 1

(257) has 2

(264) has 2

(365) has 3

(374) has 3

If we negate the four permutations that do not include ui,

then negate the resultant permutations that do not include

uj, the two permutations that include uk but not ui or uj

will be negated twice therefore will remain unchanged. The

permutation (ijk) is not negated at all. The two

permutations that include ui but are not (ijk) are negated

once, and the two without ui or uk are also negated once.

The net result is that for a defined permutation (ijk),

negating the four permutations that do not include ui

followed by negating the four permutations that do not

include uj is equivalent to negating the four permutations

that do not include uk. Clearly this may be repeated as

many times as one may like with the final result identical

to a single negation of all permutations that do not

include a single unit. If you have guessed the operation

of negating the four permutations that do not include one

of the seven vector basis units is an algebraic

isomorphism, you would be correct as will be shown.

Now look at negating the three permutations that include

basis unit ui followed by negating the three permutations

that include basis unit uj. The two permutations that

include ui that are not (ijk) are negated once. (ijk) is

negated twice so remains unchanged. In the set of four

permutations that do not include ui, the two with uj and

not uk are negated once, and the two with uk and not uj

are not negated at all. The net result once again is for

defined permutation (ijk) negating the three permutations

that include ui followed by negating the three permutations

that include uj is equivalent to negating the four

permutations that do not include uk.

To close out double negations, lets do three/four. Order

does not matter here. After negation of the three

permutations that include ui, negate the four permutations

that do not include uj. The two permutations including ui

that are not (ijk) are negated twice so remain unchanged.

The two permutations that include uj but not ui or uk are

not ever negated, and the two that include uk but not ui or

uj are negated once, and (ijk) is negated once. The net

result is that any four/without and three/with negation

combination for ui and uj is equivalent to negating the

three permutations that include uk.

If you have guessed the negation of the three permutations

that include one of the seven basis units is not an

algebraic isomorphism, you get a gold star. In fact, the

two negation schemes above are the ONLY schemes that will

result in a valid Octonion Algebra.

Any member of Left or Right Octonion Algebra is related to

one of the seven others of the same type by the negation of

all permutations that do not include one of the seven basis

units, one to one. The Left/Right cross type negation

scheme for each of the sixteen to one of eight of the

other type is either the negation of all seven

permutations, or the negation of the three permutations

that include one of the seven basis units; one to one of

seven.

For those who loathed word problems, some specific negation

examples might help to visualize this.

original not u1 not u2

(123) (123) (123) unchanged, has u3

(761) (761) -> (671) changed, no u3

(572) -> (752) (752) changed, no u3

(653) -> (563) -> (653) unchanged, has u3

(541) (541) -> (451) changed, no u3

(642) -> (462) (462) changed, no u3

(743) -> (473) -> (743) unchanged, has u3

original with u1 with u2

(123) -> (213) -> (123) unchanged, has u3

(761) -> (671) (671) changed, no u3

(572) (572) -> (752) changed, no u3

(653) (653) (653) unchanged, has u3

(541) -> (451) (451) changed, no u3

(642) (642) -> (462) changed, no u3

(743) (743) (743) unchanged, has u3

original with u1 not u2

(123) -> (213) (213) changed, has u3

(761) -> (671) -> (761) unchanged, no u3

(572) (572) (572) unchanged, no u3

(653) (653) -> (563) changed, has u3

(541) -> (451) -> (541) unchanged, no u3

(642) (642) (642) unchanged, no u3

(743) (743) -> (473) changed, has u3

original not u1 with u2

(123) (123) -> (213) changed, has u3

(761) (761) (761) unchanged, no u3

(572) -> (752) -> (572) unchanged, no u3

(653) -> (563) (563) changed, has u3

(541) (541) (541) unchanged, no u3

(642) -> (462) -> (642) unchanged, no u3

(743) -> (473) (473) changed, has u3

Notice how the union of arrows on the top two and likewise

on the bottom two fills in all positions.

Lets finish up on the algebraic isomorphism idea. Working

out the four specific examples we just did, my claim is

that the top two are algebraic isomorphisms, and the bottom

two are not. The top two are equivalent to a single

negation of all permutations that do not contain u3. If we

gather up the three unchanged permutations with u3 central,

pick any two permutations, exchange between the left two

units and exchange between the right two units, the three

are recovered:

Try exchanges 2 <-> 5, and 1 <-> 6

(231) -> (536)

(536) -> (231)

(437) -> (437)

Continuing the same exchanges on the remaining four:

(671) -> (176)

(752) -> (725)

(451) -> (426)

(462) -> (415)

These exchanges have restored the original permutation set.

Exchanges carried out consistently on all seven

permutations are nothing more than a consistent renaming of

the units, so the difference between the two

representations is basis unit naming convention only. The

algebras are indeed isomorphic.

Now for the proposed non algebraic isomorphisms. Again,

grab the three permutations containing u3, with u3 central:

original final

(231) (132)

(536) (635)

(437) (734)

Grab another three permutations with some other unit in

common, say u5:

original final

(257) (257)

(653) (356)

(154) (154)

For the u3 common triples, 1, 6 and 7 remained on a common

side, yet the side switched as required for the move from

Right to Left Octonion Algebra. For the u5 common triples,

the side for another permutation changed as expected, again

from Right to Left, but the permutation changed from

(734) to (231).

It might be sufficient to some to already see that the

consistent side swap for one of the other permutations

resulting from a change between Right and Left Octonion

Algebras would unlikely be remapped back to the original

side by a consistent name exchange for any combination of

units. If more is needed, lets try the unit exchange

method.

For u3 common, the exchange map back by unit name exchange

is 2 <-> 1, 5 <-> 6, and 4 <-> 7. Carrying this out on the

remaining four permutations:

(761) -> (452)

(572) -> (641)

(541) -> (672)

(642) -> (571)

For the u5 common, the exchange map back is simply

3 <-> 6. Carrying this out on the remaining four

permutations:

(213) -> (216)

(761) -> (731)

(642) -> (342)

(473) -> (476)

In both cases, the remaining four permutations exchange to

new triplets not found in the original. The unit exchange

mechanism fails. The algebra formed by negating the three

permutations which include any single unit is not

isomorphic to the original algebra.

The three permutations that include one of the basis units

can always be used to determine the unit naming map between

any other isomorphic representation, even those with quite

different triplet choices. There are many different

mappings between any two. Take for instance the algebra

created by starting with (124) and adding one modulo 8

omitting 0 to each unit to form the permutation set

(124)

(235)

(346)

(457)

(561)

(672)

(713)

First, identify if this is Right or Left Octonion by

examining the three permutations that include your choice

of unit number. I will pick u2. Arranging we have

(124)

(523)

(726)

Units u4, u3 and u6 are members of one of our seven

permutations so we have Right Octonion Algebra. Notice none

of the triplets match any of those in the Right algebra at

the top of this discussion. If we grab any arbitrary unit

in the first algebra, say unit u1, then place the three

permutations including u1 in arbitrary order we have a one

to one mapping between these two representations of Right

Octonion Algebra.

(617) <-> (124)

(312) <-> (523)

(415) <-> (726)

This is equivalent to

6 <-> 1

1 <-> 2

7 <-> 4

3 <-> 5

2 <-> 3

4 <-> 7

5 <-> 6

Performing the left to right map on our original algebra

(123) -> (235)

(761) -> (412)

(572) -> (643) -> (346)

(653) -> (165) -> (561)

(541) -> (672)

(642) -> (173) -> (713)

(743) -> (475) -> (457)

The second change is the isomorphic map negating all

permutations that do not include u2. This gets us to the

second form of Right algebra.

It will be quite worthwhile to look closely at the

isomorphic negation scheme. There are two triplets of units

assignable to every vector basis unit. One triplet is a

permutation and one is not. For instance, take unit u4 and

arrange the seven Right algebra permutations as follows

(123) (761) (541)

(572) (642)

(653) (743)

(123) is the "fourth" permutation associated with basis

unit u4. u5, u6 and u7 are the non permutation basis units

associated with basis unit u4. The three permutations that

include u4 give the one:one connections u1:u5, u2:u6 and

u3:u7. Now make an association between these two triplets

and our familiar rectangular physical x, y and z as follows

{xyz} : {123} : {567}

Permutation (123) covers {xyz} with a closed set rule for

multiplication. {567} covers {xyz} with an open set rule

for multiplication requiring three permutations that each

include one unit of (123). Notice that (123) is an {xyz}

right hand rule and {567} is an {xyz} left hand rule.

When the four permutations that do not include u4 are

negated in the Octonion algebraic isomorphism, the {xyz}

opposite hand relationship is preserved with (123) and

{567} both changing. This is the case for both Right and

Left Octonion Algebras. The unit exchange scheme that above

was successful in recovering the original permutation unit

orders above will be {5 <-> 6 and 1 <-> 2}, or {5 <-> 7 and

1 <-> 3} or {6 <-> 7 and 2 <-> 3} here with unit u4 instead

of u3 used in the above example. Any of these exchanges can

be seen to be an {xyz} hand rule change for both (123) and

{567} if the {xyz} connection to both remains fixed. This

effectively undoing the negation of the four permutations

that do not include u4 through renaming.

The variability in {xyz} handedness is non-confrontational

to the concept of Octonion algebraic isomorphism. The

variability only arises when an extra-algebraic connection

is fixed between algebraic units and non-algebraic {xyz}.

These geometric assignments are of no fundamental concern

to the algebra itself, even though we will need to make an

association in order to apply Octonion Algebra to physical

reality.

In conclusion on the subject of the full cover of the

definition of Octonion Algebra, there are sixteen different

juxtapositions of units within the seven permutations

central to the definition of the algebra. The set of

sixteen is actually two sets of eight, defining eight Left

Octonion Algebras, and eight Right Octonion Algebras. All

eight in each type are isomorphic algebras. However, Left

Octonion Algebra is not isomorphic to Right Octonion

Algebra.

The negation of the four permutations that do not include

one of the Octonion basis units is an isomorphism of

algebras. The negation of the three permutations that do

include one of the Octonion basis units is not an

isomorphism of algebras.

The only valid permutation negation schemes are the

negation of all three permutations that include one of the

Octonion basis units, the negation of all four permutations

that do not include one of the Octonion basis units, or

multiple combinations of these two. Any other scheme will

not produce a valid Octonion Algebra. All combinations of

valid negations will produce a representation found in the

group of sixteen.

This last paragraph is extremely important to discussions

to come when I will discuss the concept of algebraic

invariance and variance.

What follows is an itemization of the 16 different

permutation configurations for Left and Right Octonion

Algebra. Taking column 0 as the prototype, column n not 0

of the same type is column 0 changed by negating the

permutations that do not include basis unit n. The map

between types is same column negation of the permutations

that include the basis unit u4.

Left O Algebra

Column 0 1 2 3 4 5 6 7

(123) (123) (123) (123) (321) (321) (321) (321)

(761) (761) (167) (167) (167) (167) (761) (761)

(572) (275) (572) (275) (275) (572) (275) (572)

(653) (356) (356) (653) (356) (653) (653) (356)

(145) (145) (541) (541) (145) (145) (541) (541)

(246) (642) (246) (642) (246) (642) (246) (642)

(347) (743) (743) (347) (347) (743) (743) (347)

Right O Algebra

Column 0 1 2 3 4 5 6 7

(123) (123) (123) (123) (321) (321) (321) (321)

(761) (761) (167) (167) (167) (167) (761) (761)

(572) (275) (572) (275) (275) (572) (275) (572)

(653) (356) (356) (653) (356) (653) (653) (356)

(541) (541) (145) (145) (541) (541) (145) (145)

(642) (246) (642) (246) (642) (246) (642) (246)

(743) (347) (347) (743) (743) (347) (347) (743)

Rick Lockyer

For more information see

http://www.octospace.com/files/Octonion_Algebra_and_its_Connection_to_Physics.pdf

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