Metamath
|
Modelling the Holy Trinity.
So, discussion of infinite regression aside, the regression of God's
own spatial perspective, that what is outside his person is yet a fully
realised part of his own understanding including his awareness of
himself in that space and his understanding in understanding of himself,
etc etc.
Imagine closing your eyes and still being in perfect understanding of everything around you - including behind you, the colour of it all, its presence, and motion. Now imagine it arranging itself to make things more pleasing to you - open your eyes, has it changed and become as perfect? No? Well, you are not God! (You would be insane to think you could do all this otherwise. If you still think you may, you are insane.)
I have mentioned that there is an almost "doughnut shape" symmetry of
repeated congruence. A simple model is to label the 'self' of God as
'odd' and his perceived spatial surroundings as 'even'. Two unbounded torii interlock each other flawlessly. So, I propose
modelling this with the field Z / 2Z. Now, so that God may conceive of
himself as greater, by conceiving his own existence outside of his own
understanding, we require a second God, (i.e. God's Son, Jesus) so that
God may know as we do His existence as through Anselm's argument. The Bible talks of the
Son glorifying the Father, and this is how I think it works.
Of course we require that God completely contains all understanding of his Son and the Son
in some like manner contains his Father. If we were to see them side by
side, standing shoulder to shoulder we could without loss of generality
consider one of the pair as the 'even' part of the other's 'odd'. I find
this view flawed however, since there is no absolute measure of the
spatial awareness of the pair shared. Technically, The outside
world would be seen by the Father as his Son's mind, and vice versa, so
where then could they stand side by side with awareness of some three
dimensional space, a container, for them both.
An intuitive step is to use the additive group of Z / 3Z, so there is a
zero element, and the two elements 'one' and 'two'. At least in this
model we have our Father and Son, which are both order three elements,
(a + a + a = 0); yet again, I find this flawed, because both Gods,
though able to conceive their individual existence outside of their own
understanding, in the sense of the 'zero element' world, (and indeed the other
God.) both God's must be a container for all of the elements, {0, 1, 2}
in order to be "conceived as greater than any other.". Introducing some
permutation of this set for '1' and '2' based on the elements themselves
as permutation 3-cycles, the Father ≅ (1,2,0), the Son ≅ (2,1,0), and 0
≅ e, how could '0' contain both the Father and Son? Simply by being the
product of the two, it loses it's own merit as a container for two gods,
as well as yet being the sole place that God, Father and Son can
share their conceived necessary existence, and Gods should always agree
flawlessly. Thus I require a third member of the godhead, (Holy Spirit)
That said, The pair need as much a third God to glorify each other as
they need themselves. That is, they may agree in pairs of a Trinity, and
be in agreement. Symmetry must apply in that every element either takes
the form of any of 0, 1 or 2, which is not pretty, or we need some other
model.
Well, back to the first model of Z / 2Z. Why not simply extend it?
there was nothing wrong with the ability of God to perceive himself in
'even/odd', so why would we alter this to Z / 3Z where we find the
Godhead unable to perceive himself as an individual?
Well, simply put, I choose GF(2, 2) (GF for "Galois Field") of order four as my model for
the Trinity. The additive group has a symmetry to it unlike any other,
and is called the Klein four-group. the product of any two non-identity
elements is the remaining third. Non-zero elements are order two, so x+x=0
in the additive group. The multiplicative group is cyclic order three as
Z / 3Z. However, unlike a finite field, I would choose not to label the
elements as powers of x multiplicatively in a one to one assignment with
the non-zero additive elements, In order to preserve the symmetry of
each member of the Godhead. The consequence is a floating unity.
Yet, though the Klein four group is a good model, it does need some
tweaking. Each God needs by definition to be a container of all the three,
and the '0' of 'existent background' itself. We have one element
already, Gal(2 , 2). We require another, simple really, we use Gal(2 ,
3) the field of order eight. The former is not a subfield
of the latter, However, the additive group of Gal(2 , 2) is a subgroup
of the additive group of Gal(2 , 3). (Often called the Octal
group).
Now, here is the true beauty of the problem. The additive elements
and their subgroups exist... We may permute the elements of the
octal group by the multiplicative group of Gal(2, 3). So, why not the
additive subgroups?
If the elements of the octal group are (0, a, b, a+b, c, a+c, b+c,
a+b+c) and it's subgroups are;
(0, a, b, a+b)...(0, a, c, a+c)...(0, a, b+c, a+b+c)...(0, b, c, b+c)...(0,
b, a+c, a+b+c)...(0, a+b, a+c, b+c)...(0, c, a+b, a+b+c).
Perhaps, not such a coincidence that there are seven subgroups.
From a comparison with finite fields it can be shown that a generator
of the multiplicative group of gal(2 , 3) in this case say g is the
permutation cycle g=(a,b,c,a+c,a+b+c,a+b,b+c) as such an element,
written in the order of permutation of the individual elements. However,
it is very special in that it also permutes the non-zero elements of the
additive subgroups as if they were elements!
So g permutes the subgroups as..
(0, a, b, a+b) => (0, b, c, b+c) => (0, a, c, a+c) => (0, b, a+c, a+b+c) => (0,
c, a+b, a+b+c) => (0, a+b, a+c, b+c) => (0, a, b+c, a+b+c)
Of course, all powers of the element g also work! (NB: the
seventh power, is 'e' (or unity) and therefore holds the elements
unchanged.) I have spent much time
scratching my head and breaking my pencil, and I found that there are,
for any particular octal group, 48 seven cycles that work in each case,
(8 separate groups).
Of course these fall in to sets of six for the multiplicative groups of
Gal(2 , 3). (Unity is not a seven cycle. and indeed unity may be a
different element from the additive group in each case. This leads me to believe that since only
seven elements may be unity, having eight separate groups of order seven
is a strange occurrence).
Continue To Next Page
Return To Section Start |