Existence never really began. The term, "I think, therefore I am" is "Cogito, ergo sum." in latin and in describing reality we say "Sum, ergo sum." The universe simply is. We begin with a point, the zeroth dimension. We move it vertically to create our line, now we can rotate it 360 degrees vertically to create the ultimate two dimensional geometry, the circle, which has more area than any other shape. Now, we can spin this circle horizontally 360 degrees to create the ultimate three dimensional structure, the sphere, which is a shape with infinite angles and the least possible volume of any geometry. This is oppossed to flat space. Now, it can be everted as Smale & co. have shown. Now place a 3 dimensional sphere on top of it's everted counterpart, and we have 6 dimensions. However, we must account for time, which is the circle going through all possible deformations. Now, we can see time as 2 dimensional, being positive and negative in a lateral continuum, this fits with multi-variable 2 gradient vector calculus used in this model. Therefore there is a basic of 8 dimensions in which infinite more dimensions arise of time. 
Due to the spherical nature of reality, Pi is an integral part of my work as well. My work has everythibng a mathematical proof needs, inegral and ifferential calculus (scalar metric proofs) matching Hawking's equations for black hole evaporation, and a rank 12+ vector calculus for determining the trajectories, polarities, and the production of particle pairs including dark matter. All you have to do is define x1 and y1 to get z1, then y1 and z1 to get x2, and x1 and z1 for y1, and finally x2 and y2 for z2. You can find each of the infinite points on the surface of a sphere using that method. Everything in the observable universe can be collated in casaul relation. 
V(c)=x/((2(3^59))=1.616229e-35
x=1.616229e-35*((2(3^59)))
x=4.56758796e-7
4.56758796e-7/(2(3^59))=1.616229e-35
1.616229e-35=squareroot((1.0545718e-34*6.67408e-11)/c^3)
2.6121962e-70=(1.0545718e-34*6.67408e-11)/c^3
c^3=(1.0545718e-34*6.67408e-11)/2.6121962e-70
c=cuberoot((1.0545718e-34*6.67408e-11)/2.6121962e-70)
c=299792379.714
Now we suppose matter is trapped light. You can fit 8 spheres around the surface of a sphere, if these 9 spheres
represent a charged particle with 9 times the mass in 1/9th the volume of
your original photon than you can repeatedly perform these 9-fold compressions 28 more times before you exceed the planck mass. I calculated
that the entire 29th sphere would be <lp:
Since it’s charge must compress the photon by 1 planck length per planck
time for it to travel at c, the photon mass can be expressed by the quotient of
radii between a photon & planck length -> (7e-7/2)/(2(3^60))=4.1282194e-
36. Viz a viz, the photon density of elementary particle (EP) 1 is
4.9320464e-36/(4/3pi(7e-7)^3)=3.388006e-17 kg/m^3. Ergo, the particle
density of EP 28 is 4.9320464e-36 x 9^28/(4/3pi(1.6e-
35)^3)=1.4848022e+96 kg/m^3 this pretty much checks out as the densest
possible EP before you get a black hole planck particle.
We've not accounted for the interior of a black hole. Let's say there is a maximum density before the EM interaction can absorb zero radiation, ergo the EH. It forms a force field, a spherical, hollow force field whose volume depends upon the amount of spacetime getting crunched at once.
If it is hollow than information in the force field can slip through given it moves beneath the planck scale. Filling the void, that's where the code starts anew - where the fractal recurs.
Consider there's somewhere between 2 & 3 real physical dimensions at any given
point in space and time per brane, so for two parallel spacetimes of opposite linearity (torsion):
6>n>4; n=(4,6)
f(n)=(λmax)•((4π/3)r^3)
f(x)=6/n/(4π/3)^(1/3)) where n>6
f(x)=4/(n/(4π/3)^(1/3)) where 4>n
The size of an antiproton is 10^-15 m and the Schwarzchild radius of its central black hole should equal the rate at which black holes evaporate.
The Schwarzchild radius is 2.484e-54 meters (just type proton into where it
says earth). The rate of evaporation is 8.41e-17 seconds (just type proton
into where it says earth).
But protons do not have λmax = vacuum density, that’s the problem, so for a
proton we must use the original equation f(n)=(λmax)•((4π/3)r^3); where
f(x)=4/(n/(4π/3)^(1/3)) where 4>n to find the contraction of c with the λmax
of a proton ≈ 395 nm. However, in the special case of black holes the
equation must be modified.
First of all, it’s 4πr^2 because the quasar within the Schwarzschild radius of
the antiproton is a hollow sphere. Secondly, λmax of the proton’s collective
micro-BH quasars is the proton’s normal λmax but to the negative power of
the proton’s length divided by twice the Schwarzschild radius
f(n)=(3.95e-7^-(1e-15/2(2.484e-54)))((4π)(2.484e-54)^2)=7.753772e-107
f(x)=4/(7.753772e-107/(4π))^(1/2) = 1.610306e+54
So a black hole with the mass of the sun (1391400000 meters) has a
Schwarzschild radius of 2953 meters & will evaporate in 6.61e+74 seconds.
f(n)=(5.04e-7^-1(1.3914e+9/5906)) x ((4π x 2953)^3) = 2.3886249e+25
f(x)=6/(4π(2.3886249e+25^(1/2))=9.7693891e-14
1.610306e+54/299,792,458/9.7693891e-14=5.4981971e+58
5.4981971e+58/8.41e-17=6.5376898e+74 seconds ✓
So we rescale the photon after the final transformation.
v(QE)=X/(2(3^59))=1.616229e-35/(9^28)
X=(2(3^59))*(3.0882513e-62)
X=8.7276366e-34
8.7276366e-34/(2(3^59))=3.0882513e-62
3.0882513e-62=squareroot((1.0545718e-34*6.67408e-11)/c^3)
9.537296e-124=(1.0545718e-34*6.67408e-11)/c^3
c^3=(1.0545718e-34*6.67408e-11)/9.537296e-124
c=cuberoot((1.0545718e-34*6.67408e-11)/9.537296e-124)
c=1.946917e+26 m/s
What is indicated is that we live inside one such black hole nestled in the everting spacetime. This gives us the mechanism for all motion that occured before any of the planck particles formed.
The horizon is about 14 billion light years away, that's where we see the picture of the Cosmic Microwave Background at the edge of the quasar around our black hole. Our black hole was built from planck particles in a super cosmos that were about about 7e-7 meters across which merged instantaneously in the core of a very large collapsing star.
This black hole would be expanding from our viewpoint (remember 3 dimensions in which time flows one way) within it because all points are being dragged through their centers, sort of dragging the spacetime curtain beyond our cosmic event horizon (where the spacetime flows the opposite direction like a parallel spacetime weave), and we're experiencing expansion at the other side of the centers.
1.616229e-35x9^28=8.45849622e-8
8.45849622e-8^2=7.1546158e-15
Cuberoot((1.0545718e-34x6.67408e-11)/(7.1546158e-15)=9.9455105e-11
Square root((1.0545718e-34x6.67408e-11)/9.9455105e-11^5)=850.48386809
Supra-cosmic lp = 8.45849622e-8 m
Supra-cosmic tp = 850.48386809 sec
Supracosmic c = 9.9455105e-11 m/s
Supracosmic time dilation = 850.48386809/5.39e-44 = 1.5778921e+46 times
If CMBR is the interior of a supra cosmic eh instead of a central region expanded from than 14 billion ly is a radius not a diameter, with the "Great Attractor" at the center adding hundreds of millions of light years to the 13.8 billion light year radius. And that diameter is 28-29 billion light years.
Evap rate of about 1.5778921e+46 x 2.09700e+67 = 3.30884e+113 years for a supracosmic solar mass black hole which is 5.8 km in normal cosmic diameter. Meaning bh is 1/10th its former mass, meaning originally it's evaporation rate would have been 28,000,000,000 light years = 28,000,000,000 x 9.461e+15 = 2.64908e+26/5800m=4.5673793e+22 years as m over m° before being cubed in the bh evap equation 9.5279888e+67 x 2.09700e+67=1.998019e+135 year evap rate!
Which puts the universe black hole at
1.998019e+135/1.57789e+46=1.26626e+89
1.26626e+89/2.09700e+67=6.0384359e+21
Cuberoot(6.0384359e+21)=18209924.8488 solar masses...
We are within an SMBH in the supracosmos.
1 sphere
x^2+y^2+z^2-8x+6y+4z-7=0
4+4+z^2-32+24+4z+7=0
1+1+z^2-8+6+4-7=0
z^2+4z=7
z(z+4)=7
2+z^2=5
z^2=3
z=squareroot(3)
(x,y,z)
(-x,-y,-z)
(x,y,0)
(-x,y,0)
(x,-y,0)
(-x,-y,0)
(0,z,0)
(0,-z,0)
(z,0,0)
(-z,0,0)
Next x points
x^2+1.25+1.25-8(x)+6(1.25)+4(1.25)-7=0
We can rotate the spheres as well to view them from in the 3rd dimension from different angles:

The condensation of space (a spherical coordinate system with a sub Planck geodesic) has a 2 dimensional vector gradient equal to (22.5, 112.5) clockwise from y. The more coordinates in the system (if the volumetric geodesic is the Planck length there's 10 point particle coordinates per spherical coordinate.
f(y,-x)=[0101101101]
f(x,y)=[101]
So,
t=1
Vector 4 or f(y,-x): [-9/10lp,+9/10lp,-9/10lp,+9/10lp,+9/10lp,-9/10lp,+9/10lp,+9/10lp,-9/10lp,+9/10lp]
Vector 5 or f(x,y): [+4.5/5lp,-4.5/5lp,+4.5/5lp]
That is a ratio of 9/10 Planck lengths for the first Planck time. All of the following Planck times are the same but for t=2 we have 8/10 Planck lengths. This length contraction leads to black holes when the ratio of the coordinate system is equivalent 1/9^28 Planck lengths.

I have found my own 4 fundamental forces of nature:
1. The Planck Charge
2. Time Dilation / Length Contraction
3. Particle Pair Production
4. The Transplanckian Phase Shift/wave particle duality
No gravity, particle pairs have new patterns of Planck charge because they are born of the same generation when denser older charges become more diffuse leaving room for newer faster generations of propagating strings that slow down and collect into electrons (string intersections of larger iterations than Planck scale) or nucleic quark seas. You could say the transplanckian phase shift has to do with temporal acceleration when collections of hyper-slow string intersections create bottom strange quark star esque nucleic bodies that exceed the Planck mass.
It's not that the universe is spinning or objects are falling, it's that younger parts of the universe are adopting new angles relative to the older parts of it.