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 Metamath The Twelve Tribes The 144 octal reversals correspond to the six floating unity choices (not in the inner set of the underlying unity of the GF(8) base field that sets the 24 elders in "rest"). The (other) six choices have 24 such reversals each. Alternatively we may choose six choices of unity in eight possible C7 groups under three automorphisms of GF(8). For each choice of unity (a = 1 say) in the octal we have 4 choices of static subgroup and then 3 further choices of the corresponding subgroup that is unity in the particular K4 filter we examine (say one of, [a,b,c],[a,d,e],[a,f,g]). This immediately gives 12 possible K4 groups for making our filter from a fixed unity with the underlying octal. Now,.. given this direct association from the three of a = 1 = [a,b,c] say, we have two choices from b and c from [a,d,e] and [a,f,g]. There is something else going on here though as this choice appears more or less arbitrary to the K4 filter. We know that our K4 filter must contain itself as an element,.. and without loss we may assume it is unity. We must assume that out twelve tribes have attained a "rest state" of unity in the K4 ultrafilter, (corresponding to its own unity.) Now we can shift the unity element of our K4 filter with the frobenius map on the field GF(8). (sending b->d->f and c->e->g would make the filter a = 1 = [a,d,e], d = [a,f,g], e = [a,b,c]) Then we have three possible automorphisms of GF(8) imbuing the K4 filter with "spirit" or life (anima). Whilst we also have two possible C7's we can use to underpin each of our possible K4 filters, (giving us two different sets of three automorphisms, 6 in total.) Frobenius in the octal merely provides multiplication for the K4 filter. So we appear to have 12 * 2 so far - we desire another factor of six! We simply note that now we may introduce the choice of unity in the octal remaining apart from the 24 elders. I.e. b = 1 = [b,d,f] etc then we have accounted for all 144 arrangements of the octal. This last factor of six depends only on the choice of unity, not on the seven cycle being used, and the choice of two possible seven cycles does not appear to make a difference to the K4 filter, as the K4 multiplication used is dependent on the octal's own static subgroup and the choice of unity: and not on the seven cycle. So how do we interpret this? We see that every floating unity element that has attained rest in the Father (the GF(8) ultrafilter) can be accounted for as corresponding to the multiplicative identity of some field GF(4) (whose multiplicative group is induced under the frobenius map on GF(8).) Further, the additive group GF(4)+ is in direct correspondence to a static subgroup of GF(8)+, by virtue of alignment to a left handed group held static: leaving only one subgroup in the right hand with which we may associate the K4 filter. I.e. a = 1 = [a,b,c],[a,d,e],[a,f,g] corresponds to [c,e,g] in the left hand which only leaves [b,d,f] in the right for the group a = 1 = [a,b,c],[a,d,e],[a,f,g]. (Christ is sat "at the right hand" truly!) All 144 states that do not correspond to a fixed unity in the GF(8) field (as the 24 elders) but to its floating unities instead correspond to one of these K4 arrangements over the octal. (Allowing the static subgroups in GF(8) free floating by choice also.) The 24 elders, or identity arrangements of the GF(8) ultrafilter are separate and are not placed in these corresponding 144 arrangements of the K4 filter given that it is assumed that the Son is sat at the right hand of God and thereby directly related to the static subgroup by relation to the coresponding left hand group only, and in the octal the static group must not contain unity, (and can not be the same subgroup that is static in the K4 filter.) We assumed that the K4 filter was constructed of the three subgroups containing the unity element, and not the fixed subgroup of the octal under frobenius. In this sense it is impossible once the K4 is 'sat' at the group of these three elements) for the Son to be considered to contain the 24 elders (as [b,d,f] does not contain a = 1) - but the 'sitting' K4 filter does indeed account for all of the 144 other arrangements that do not have this unity to unity correspondence in the octal. The K4 has a "vector" of separation from the unity to subgroup correspondence of the Father's filter. It associates a different subgroup with unity. The "vector" or "displacement" takes form in the two seven cycles that preserve that K4 filter that may map the K4 filters own static subgroup to that of the octal (i.e. to the 24 elders), this pair of seven cycles relate the 12 arrangements of the K4 filter to the 24 elders' "rest state" as made over every choice of unity from the octal. The "healing" to "rest" by the operation of the Holy Spirit is in view here. Conversely, were the K4 filter to be paired as the static subgroup to unity in the octal, effectively unity to unity (in the octal) and to contain the 24 elders as a matter of course, we would certainly require an association with the K4 filter through the complement subgroups from the left handed octal (referred to elsewhere as a "hail" column) as in alignment with the right handed octal's singleton elements. In this fashion we have the K4 filter floating free from the right hand of God, and truly free to correspond to both left and right hand parts; I.e. e = [a,b,c], c = [a,f,g], g = [a,d,e] with a = 1 with {c,e,g} a static subgroup in the left hand. (And directly paired with [b,d,f] - as static in the right.) Sat at the right, there are 144 rest states across all plausible K4 filters - aside from the 24 elders. Freeing ourselves to float the K4 filter to unity at the right hand we observe the properties of the static subgroup (as under frobenius) move us to using only the complement subgroups in the left handed octal for that particular seven cycle used. (We then seek an alignment rather than a reversal). These are the prerequisite requirements of the first of the four opened seals, and the page "An Application" in the "Wise Virgins" pages of the metamath section can help explain further.Continue To Next Page Return To Section Start Return To Previous Page