# Why Octonions 2: A Full Description of the Algebra

31 views

### Rick

Aug 9, 2008, 12:04:00â€¯AM8/9/08
to
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.

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
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

www.octospace.com

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