and also 2003 Nature magazine 425 593.

Now the Luminet team arrived at their Universe shaped as a
dodecahedron from studies
of the Cosmic Microwave Background CMB, that the plot or graph implies
a dodecahedron.

So I wonder where the error of 2X or 1/2, where the error arose?
Because a Plutonium
Atom Totality would have the shape as a cube or 6 sided figure not a
12 sided figure.
Unless I was a scientist working on that Luminet team, it would be
difficult to spot the
error and may even be difficult if I were a member having done the
work itself.

So I have been looking into the assumptions of the Luminet team and
the best that I
could muster is the assumption that the CMB of the Southern Hemisphere
and of
the Northern Hemisphere are distinct mappings, whereas in the Atom
Totality, the
southern and northern hemisphere join into one since we have
circumnavigated the
Cosmos.

So, this is a guess on my part, that perhaps the Luminet error of
getting a Dodecahedron
when they should have ended up with a Cube, is that they are counting
2X the CMB when
it needed to be counted only once.

When the Johns Hopkins scientist first announced the Universe is a
color of pale green and
after shown they missed a correction factor that would make the true
color of a silvery white.
So maybe the Luminet team had some technical factor in error? Or maybe
it is the counting
of the Northern and Southern hemispheres when only one should have
been counted.

A dodecahedron is very much different from a cube, and I strongly
suspect that a 1/2 error
occurred.

Now I am looking for some other evidence that would also support a
Cube rather than
a Dodecahedron.

(1) Evidence that EM is 90 degree angle that equates electricity to
magnetism and not
a 36 degree angle of a dodecahedron

(2) I am looking at the combination of pi and e. If the Cosmos was a
dodecahedron, does
a nesting of pentagon and circle or dodecahedron and sphere, does a
nesting relate to
pi and e? Whereas a nesting of cube and sphere or square and circle
have a direct relationship of
pi and e? You see, Plutonium Atom Totality creates pi with a value of
22 subshells in 7 shells
and e as 19 occupied subshells in 7 shells. So that there must be a
favoritism for either
the Dodecahedron or Cube but not both.

So far it looks as though the Cube is the winner, but we need to
pinpoint where the 1/2
error crept in.

plutonium.archime...@gmail.com skrev:

And as there are 7 shells, I think there will be 1+2+3+4+5+6+7=28
subshells,
all of these are not shown in the table.

Alright, and thanks for the details so that my response can be
quick and easy.

Notice that the d and f in the sixth or O domain and yet the occupied
s orbital of the seventh or P domain. So I count the d and f of the
sixth
domain as "occupied".

What justification do I have for that? I say it is Quantum Mechanics
that
if if the electron structure has any representatives in a domain than
all
prior domains are "occupied". It is the Double Slit Experiment where
the
particles break apart to occupy the screen as a wave and so you have
to
include the d and f of the sixth domain as occupied.

Maybe some one else would give you a more satisfying answer than my
justification.

When you look at Quantum Physics and say that plutonium has 22
subshells
in 7 shells and has 19 occupied at any one time, then the mistakes
that are often
made are those in which people expect clearcut 100 percent black and
white answers
when as always in QM the answers are gray. The answers to the Double
Slit is that
the particles break apart at the slit and then reform at the screen
and people do not
like those types of answers (and that is why they should not be in
physics).