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 Metamath The Maker's Dozen We already noted the existence of 144 octal 'reversals' and that the identity corresponds to 24 states or "rests". For each choice of octal there are 8 different multiplicative groups preserving K4 subgroups under multiplication. There are then 7 further choices for the unity element and then three further automorphisms of each finite field GF(8). 7*8*3 = 168 = 144 + 24. We may separate out those 144 to the six non-unity elements so that there is always a direct map by the Holy Spirit's action in C7 to map each of the remaining six (as multiplicative elements or subgroups) to unity, to "attain rest". Likewise we note that under the frobenius map one K4 subgroup is held static in GF(8). For instance if a = 1, and (b,d,f) is static then with the seven cycle (a,b,d,c,f,g,e) we see; a = [b,d,f] b = [c,d,g] d = [c,e,f] c = [a,f,g] f = [b,e,g] g = [a,d,e] e = [a,b,c] So taking the subgroup for the K4 ultrafilter sat at the right hand that contains unity we have say, [a,b,c] whose three elements correspond to [[b,d,f],[c,d,g],[a,f,g]]. However reversing these with a further or subsequent octal reversal by adding a left handed octal (thereby flipping the association of the right hand list above from top to bottom) we then preserve our K4 subgroup with [[a,b,c],[a,d,e],[a,f,g]] which is indeed a subgroup formed of K4 subgroups. a = [b,d,f] = {c,e,g} b = [c,d,g] = {a,e,f} d = [c,e,f] = {a,b,g} c = [a,f,g] = {b,d,e} f = [b,e,g] = {a,c,d} g = [a,d,e] = {b,c,f} e = [a,b,c] = {d,f,g} Now note the three subgroups containing "a = 1" correspond to the elements c, e and g (and also to a left hand triple of those three, i.e. a = {c,e,g}) Now under the frobenius map 'a = 1' would be fixed, and b->d->f and c->e->g. However we have a separate correspondence for the K4 ultrafilter than the static subgroup. Since we must accept that a = 1 = [b,d,f] However we may choose the complement octal so that; a = [c,e,g] b = [a,e,f] d = [a,b,g] c = [b,d,e] f = [a,c,d] g = [b,c,f] e = [d,f,g] So now we have elements that correspond to subgroups in an alignment yet using subgroups ie [a,b,c] = [[c,e,g],[a,e,f],[b,d,e]] However for our K4 ultrafilter we simply require that 1 = a = [a,b,c], b = [a,d,e], c = [a,f,g] whilst we ignore the ultrafilters image in the octal. (In the octal the left hand alignment with a = [b,d,f] would mean we could only use e = [a,b,c], g = [a,d,e], c = [a,f,g]) We know that in the octal we have four possibilities that may correspond a subgroup to the unity element. If a = 1 then we may choose a = [b,d,f] or [b,e,g] or [c,e,f] or [c,d,g]. Then the separate K4 ultrafilter's three elements a =[a,b,c], b = [a,d,e], c = [a,f,g] may freely correspond to any of these four correspondences without affecting our unity in K4 a = 1 = [a,b,c] and its other correspondences of b = [a,d,e], c = [a,f,g]. All that is required is that we alter the particular seven cycle of the underlying octal and this does not effect the frobenius map forming our GF(4)* on [[a,b,c],[a,d,e],[a,f,g]] or change the subgroup elements a,b,c of the K4 ultrafilter. (We merely change our choice of static subgroup in the octal.) Frobenius on the octal sebds b->d->f and c->e->g which would map; a = 1 = [a,b,c] b = [a,d,e] c = [a,f,g] to a = [a,d,e] d = [a,f,g] e = [a,b,c] to a = [a,f,g] f = [a,b,c] g = [a,d,e] So frobenius on the octal induces a multiplicative group of C3 on the K4 ultrafilter. (The Father has given Christ to have life in Himself, even as the Father has life in Himself.) Likewise, we notice that the permutation group S3 on three elments has six members, {'e',(1,2),(1,3),(2,3),(1,2,3),(1,3,2)} written as permutation cycles. Then we also have similar (equivalent) K4 filters as it were under the action of frobenius on GF(4). Again, frobenius induces a cyclic group C3. a = 1 = [a,b,c] c = [a,d,e] b = [a,f,g] to a = [a,d,e] e = [a,f,g] d = [a,b,c] to a = [a,f,g] g = [a,b,c] f = [a,d,e] In these three cases we see the actions of transpositions (b,c), (d,e) and (f,g) respecively. Thus we have 4 x 6 choices (four possible correspondences to unity and six ways to choose 3 elements for K4 with a = 1, (six possible permutations of three elements)) - all our octal reversals in GF(8) correspond to a free floating choice of unity (of 7 possible states of "rest"). We should take notice of The "Sit at my right hand until I make thine enemies thy footstool." condition which would appear to fix the K4 ultrafilter to the static subgroup of the octal. In our case thus far this would be [b,d,f] corresponding to a = 1. In fact the three groups [a,b,c],[a,d,e],[a,f,g] correspond to the complement triple of [b,d,f] in the left hand, namely {c,e,g}. There is no way in the octal to map a = 1 = [a,b,c] b = [a,d,e] c = [a,f,g] to a = 1 = [a,b,c] c = [a,d,e] b = [a,f,g] Frobenius in the octal may send b->d->f whilst in the K4 form with it's multiplication we could have a->c->b, this would indeed result in the map required, but what is needed is a composite with the operation of GF(4), and not formed only from the octal. (We shall consider these two cases as automorphisms of each other under frobenius upon GF(4) and not upon GF(8).) Then we may simply consider the first three cases rather than all six permutations: (The remaining three cases are with respect to the underlying octal, an equivalent case.) Then we have choices of 3 elements * 4 correspondences = 12 : but 12 of what? If we notice the floating unity amongst these three elements of the K4 form as we had for the octal we do indeed find our answer - we have a floating unity giving three states of four choices. Taking into account the 144 possible states in the octal outside of unity (cf the 24 elders corresponding to unity of the 168 total reversals) we would appear to reduce to 2*4 = 8 possible "rest states" in K4 apart from the unity element, and not then the full 12. These "8" parts are in actuality never "rest elements" in the K4 form. In reality the frobenius map on the octal induces a 'shift' in the K4 form: For if a = 1 = [a,b,c], then b = [a,d,e] maps to d = [a,f,g] under frobenius on GF(8) etc. There is no method under frobenius in GF(8) that would map "a = 1" to "b = 1", frobenius without exception fixes the identity element.) The 24 elders correspond to the unity element in the octal, (a = 1) and in the K4 form they correspond to a = 1 = [a,b,c], the remaining 144 rest states must be related to the remaining correspondences of static subgroups (as [b,d,f] changed to [b,e,g] for example.) in the octal and not to the elements cut off in the K4 form b = 1 = [a,d,e] and c = 1 = [a,f,g] This exclusion from the fixed unity is related to the fact that the intersection of GF(4) and GF(8) is [0,1] only, so we must relate the 8 remaining arrangements of the K4 filter somehow to the 144 remaining rest states in the octal. I.e. the filter a = 1 = [a,b,c] b = [a,d,e] c = [a,f,g] maps to a = [a,d,e] e = [a,f,g] d = [a,b,c] and the parts "b,c" in the first and "e,d" in the second are "cut off" - they are not preserved under the frobenius map of GF(8) inducing the K4 forms multiplication. This reduction to a third may be related to the formula "in all the land I cut off two parts and a third I bring through the fire". The element and subgroup that corresponds to unity in the K4 filter must also correspond to the floating unity of the 168 total possible octal reversals. With unity considered fixed here, we should show that the two non-unity elements in the K4 filter are not as a "state of rest", i.e. not as 'rest' or 'unity' in the K4 filter or distributed throughout the octal filter in a sense that they correspond to any choice of floating unity. (Never being as unity unless they are transformed upon, which would change unity from 'a' as here to 'b' or some other say.) Then we see that we have twelve states in total, but we can not float the unity from 'a' to 'b' or 'c' without acting upon the underlying octal - altering the subgroup associations from a =[a,b,c], b = [a,d,e], c = [a,f,g]. to some other set without a = 1. Even reducing from 4*6 = 24 arrangements of the K4 form to the 3*4 = 12 arrangements has consequences. In the octal we find that we have 12 such sets in pairs of seven cycles (for there are two C7 groups that hold static the same unity and subgroup of the octal.) Then there are 12 states "In Christ" that may be mapped onto the rest state of the 24 elders by six cycles (elements) from each of two C7 groups. There are then "twelve tribes" in Christ and there are 2 maps by seven cycles back to "rest" (unity) for each. (I.e. that map the static subgroup (as [a,b,c]) in the K4 ultrafilter to the static subgroup (as [b,d,f]) in the octal ultrafilter.) However, even with the reduction to a third (two parts cut off in the K4 form) we still have three choices for the static subgroup in the K4 form associated to 'a', as one taken from [a,b,c] [a,d,e] and [a,f,g]. We must simply accept that for each underlying octal the "rest state" of the K4 ultrafilter is fixed aside from the other two elements in the K4 form (paired with b and c as above). In this way we preserve the K4 ultrafilter with those two alternative maps (within the octal, mapping it to its static subgroup) by appropriate seven cycles. This gives us our full 24 states per unity from the octal. Thus we can see that the 144 reversals correspond to those elements which are "In the Son" or have a K4 rest state, then we may assume they are "in the Father" with equivalence. The fact is that there are two seven cycles that hold fixed each choice of unity in the octal with a particular static subgroup. In fixing the K4 ultrafilter on that basis only, (as up to those two C7 groups) there are 12 rest states for each association in the octal that fixes the K4 ultrafilter (three choices for static subgroup in GF(4) and four choices of static subgroup in the octal). The remaining factor of 168/12 = 14 is shown by the seven possible choices for unity and the two seven cycles over the octal that are appropriate to preserving that K4 ultrafilter - and mapping it back to the static subgroup in the octal. (as [a,b,c] to [b,d,f] here.) One could make the easier assumption upon the octal that the states of the 24 elders split into like sets of 12 (as in paired C7 cycles) with the same fixed unity and static subgroup: the "earthy elements" outside the static subgroup are permuted with respect to each other, but then they are not being considered - the fixing of the K4 ultrafilter as "sat at the right hand" is actually unrelated to this permutation of those remaining four. This leaves us with 12 states, the number of the tribes of Israel, but more appropriately the number of the sealed tribes in revelation that may freely turn to the left or right: they are at rest. Continue To Next Page Return To Section Start