Walking Amongst The Candlesticks

Our construction in vectors was as previously;

LH = (GF(4)+, Octal (right handed), C7) and
RH = (GF(4)*, C7, Octal (left handed))

GF(4) = (GF(4), RH, LH)

The right hand octal is in reversal to the operation of its singletons under addition whereas the left hand octal is perfectly aligned to the right hand singletons. We may recover the same invariance we have in the right hand under multiplication (by the seven cyle) in the left hand under addition, as long as we use subgroup addition on left handed triples - they align with the operation of addition on the singletons and so in order to have invariance on the right handed subgroups, we must first translate to the left hand octal.

In this sense there is a divison between "odd" and "even", in that the left hand elements appear totally "even" and the right hand elements "odd" since right hand subgroup addtion appears to be only in alignment when we add with singletons that are members in those subgroups, and the action is to hold three subgroups static and permute the remaining four. (These four correspond to the "earth" or the coset of the K4 form of the ultrafitler taken from the set of seven K4 subgroups.)

We should now look at the invariance under the other seven C7 groups that are valid for our right handed octal. From the mystery of the seven candlesticks in revelation we have in columns a generator of seven, the top row being our familiar example from before on the usual right hand octal from elsewhere on the site.

a b d c f g e
c f a e g d b
e g c b d a f
b d e f a c g
f a b g c e d
g c f d e b a
d e g a b f c

The static K4 subgroup "walks amongst" the candlesticks in a fashion that is it regularly spaced in the seven cycles as long as "unity" floats from element to element in accompanying the static subgroup.

We expect to recover the exact same invariance that we had before. We simply need to note that any proper (non-unity) and non-trivial (not from the same group of the first row) seven cycle will generate an element of each of the other seven groups.

Under addition though we have a conundrum! for each seven cycle is only valid on two octals, and we must show that under any particular choice of seven cycle on the left hand vector above we recover the "same" right hand octal we had before: a very confusing idea, but one I will attempt to explore.

We of course, must require addition rather than multiplication. We must note that the left hand octal is "free to float" under the choice of C7 cycle common to both left and right hand octals as under the full eight C7 groups over the right hand. Then there is always a left hand octal for any choice of seven cycle (so that the C7 is common to both left and right hands), but the left hand octal is therefore not constant and must therefore float. Every possible left hand in view must fulfill the alignment to the right hand singletons as before.

Then, every left hand octal is under an isomorphism to the right hand singletons.

We could replace the singleton elements in our top row of candlesticks with the left handed triples to which they align, (for that particular seven cycle and right hand octal) but we would have to use different left hand octals for every subsequent row. We now have the "diamond bullet moment", where we require that the generating column would always be a valid seven cycle over every derived left hand in the table. The columns would be as from a constant "left hand" octal common to all the others. (A complete impossibility surely?)

In effect, we must shift position so that our generating column has a constant left hand, and not the rows themselves, which would have a different octal each one. However as a result we have to balance our vectors, so that the left hand addition corresponding to the right hand singletons of our generating column's seven cycle are so as to preserve invariance as before.

We have something almost orthogonal!

Every row would generate a different left hand to the column's commonly held seven cycle. Wheras before we only had a sense of multiplication valid for eight C7 groups, we now approach a set of eight left handed octals that are likewise related in a similar fashion but under addition!. Every left hand must be isomorphically in alignment with the common right hand octal's singletons (in the columns).

Invariance under addition then allows the seven generated left hands of the rows to preserve with addition their alignment to the singletons in each row, and our "reference (column) left hand" is in algnment with the singletons of the columns. Thus, every row may be acted upon by addition using the operation of the left handed octal of the columns under that particular right/left association derived from the column's seven cycle.

The columns would all align in the one same octal, and allow us to add the derived left hands together, in a manner that aligns to the singletons of the columns, rather than the rows. Every left hand of the eight is in alignment and is isomorphic, so we note the persistence of the static subgroup regularly placed in the rows and columns, (with a floating unity), and we happily state that the rows are in alignment to each other, therefore we have a system where we can add on the basis that we do so in subgroups. (Though we are restricted to adding them in similar columns) The result is then always the same for any left hand. The regular spacing of the static subgroups shows that the left hand alignment to singletons preserves the invariance under addition that we were hoping for.)

Now, returning to our "diamond bullet" - that the seven cycle in the columns is valid over every right hand octal in singletons in our table. Truly, we are holding one seven cycle as constant and we preserve invariance under addition with eight possible left hand octals, whereas before we only had a sense of holding the right hand octal constant varying with eight seven cycle groups to generate eight left hand octals.

Now returning to our vectors;

LH = (GF(4)+, Octal (right handed), C7) and
RH = (GF(4)*, C7, Octal (left handed))

GF(4) = (GF(4), RH, LH)

Choosing every proper non-trivial seven cycle as before, we may simultaneously act on the structures of the top two rows and likewise preserve the product of the results in the third. We have to make sure this construction is uniform as in the "columns" from our table. We may do so with any ordering of seven cycles we wish that preserve the static subgroup's regular spacing. There will always be a common right hand - and therefore the "common left hand" must be found in our columns, although we are free to choose any seven cycle (or left hand octal) for the column from the remaining seven and we happily have the same result, irrespective of whether we generate rows from our column or columns from our rows.

In effect the first component GF(4) is invariant under multiplication by any seven cycle, as is the right hand and C7 in the second components, the third is invariant in the multiplication across rows in our table because of the isomorphisms and bijections between our varying left hands.

The converse for addition must be done so in a manner common to every row and column. There is one octal common to all eight C7 groups, our right hand only. If we admit the revelations description of a "sea of glass" we would entail that the regular spacing of the static subgroup in each row and column of our table identify that the group structure is common under addition.

What we may do at this stage is translate to the K4 form of the ultrafilter and add groups of K4 subgroups together - each right handed subgroup with a=1 would contain unity, and we have no need to show that this again separates into odd and even. When we would add left handed triples we would require a common octal, and once again we would have to show an alignment - which we indeed have so through the intersection of any three K4 subgroups. Then addition in right handed group-triples is the common octal we require to all in addition, and we have a simple transformation that preserves the structure invariant over addition, this time with the component "LH" in right hand triples - not left hand triples which are free to vary! The one common right hand octal will always have the same subgroups, the left hand is isomorphic to this octal of group-triples.

(The right handed octal in "LH" is in the same K4 subgroups, the left handed octal in triples of those groups also. We now have a symmetry then.) In effect, Christ is sat at the right hand, but is standing at the left.

There is one sufficient necessary condition remaining! We must have the right handed group-triple as GF(4)+ or GF(4)* is actually unity in the right (and left) hands as required, showing us that there is an intersection between the octal and K4 forms of the ultrafilter (when unity) as required, and we are almost done.

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