Left And Right Hands As Vectors
So in each case of left and right hands we have three components. H, G and X, (X to which x belongs). Then xG < xH also.
So our three components can be stated as {H, G, X} or, in our case;
LH = (GF(4)+, Octal (right handed), C7) and
RH = (GF(4)*, C7, Octal (left handed))
I.e. in the left hand (LH) the right handed octal with its right hand subgroups are invariant under multiplication with x in C7. These additive subgroups isomorphic to GF(4)+ align with the groups of singletons in the left hand.
In the right hand (RH) the frobenius map which preserves the static subgroup GF(4)* (that arises from the multiplication of C7) is unchanged (invariant) despite the elements being permuted (in alignment with the left hand) with sums of elements from the left hand. Alternatively the static triple in the left hand with the sum 'e + g = c' corresponds to the K4 filter in which the sum [a,b,c] + [a,d,e] = [a,f,g] is in alignment. These groups are permuted in the right hand by the frobenius map to form GF(4)*.
So the left hand is invariant under multiplication in the right hand (The reversal does not apply as we use multiplication) We have in simple terms xG < xH
In the right hand multiplication induces the same K4 ultrafilter on the same three subgroups. we have X + G < x + H.
In both cases you may put G = GF(4) with multiplication on the right and addition on the left.
So were we to use some "scalar-style" product * component for component, could we imagine with C7 * Octal = GF(8)...etc
{0, (GF(4), GF(8) (right handed) GF(8) (left handed))} = {0, (self, RH, LH)}
Where our desired GF(4) ultrafilter of {0, Son, Father, Holy Spirit} is made by the associations of LH to Father, RH to Holy Spirit. (We assume that the element "self" is the GF(4) ultrafilter itself. A self-referencing condition.)
The question arises, is LH+RH additive or is + a form of multiplication in GF(4)*?
In forming the field GF(4) from the second components of LH and RH above, we must have them the same elements, should we not? At the very least there is a multiplicative element that will map the three elements in each candidate for GF(4)+ to the elements in GF(4)* of the other hand. In this manner we know that as our structures are invariant, there is always such a solution.
So (0, GF(4), RH, LH) is the case, since there is a multiplication in both the right and left hands that could map the elements of the two groups GF(4)+ to GF(4)* and vice versa from our two vectors. (There is also an addition of a singleton in the left hand that does the job if GF(4)+ is a right handed subgroup, and the case is as in RH above.) and there is a product by seven cycle in the other case, as LH also.
These two elements are unique to the separation of GF(4)+ to GF(4)*, there are seven such differences in each vector LH or RH. The invariance is simply up to choice of unity say, (without a better reference point) And the elements in GF(8) resolve in the sense of right hand or left hand to the multiplicative and additive operations that do the same job but as inverses by mapping GF(4)* to GF(4)+ and vice versa..
We have noted the "sit thou at my right hand..." principle which would shift from the correspondence in the octal ultrafilter with a = 1 of e = [a,b,c], g = [a,d,e] , c = [a,f,g] (singeltons in the left hand octal, subgroups of the right.) to the K4 form of the filter a = [a,b,c], b = [a,d,e], c = [a,f,g] by relating the subgroups to {c,e,g} and directly then to {b,d,f}. There is then a unique pair of seven cycles that will map [a,b,c] to [b,d,f] as we require.
In the vectors RH and LH above we may separate out the elements that belong to the K4 part of each vector. Clearly if x is in G then x + g is also in G (given any g in G). Harder to see but also valid is that squaring of elements by a seven cycle holding [b,d,f] static permutes the elements of [b,d,f], yet we *can not* separate out these elements as a set of three in the seven cycle belonging to K4.
The result is that the ambiguity (out failure to separate out GF(4)* elements) is only in the right hand vector RH where an additive element in the left hand octal either belongs to the GF(4)* part or the coset of the similar GF(4)+ elements in the left hand octal. In "sitting at the right hand" The product of the vectors 'RH + LH' results in the field GF(4)'s elements as substructures of the right handed GF(8) under both the multiplication and addition of the left hand GF(8) (and is totally invariant using this order, as under operation of the left hand GF(8)), as the same multiplication works for 'both' hands and left hand addition works correctly in alignment with that of K4 subgroups of the right hand octal.
The result then is that the RH+LH = GF(4) field i.e. (0, GF(4), GF(8), GF(8)) may be considered to lapse into the arrangement of vectors (0, K4, RH, LH) when "sat" at the right hand. If GF(4) were sat at the left we could exchange the components of RH and LH to get a similar result.
However, RH represents the structure of Christ contained in the Holy Spirit (C7) invariant under operations in the Father, and LH the states of Christ as K4 subgroups within the Father, invariant under operations of the Holy Spirit.
The product RH+LH then represents the Father in the Son under operations of the Holy Spirit and The Holy Spirit in the Son under operations of the Father. I think that was what was wanted. I hope to discover more from the Lord in the future.
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