The Trinity - A Christmas Gift

Posted on Monday, December 17, 2012 at 10:23AM by Doug

In the New Physics blog, we announced the exciting development (or latent discovery) that the ratio of the continuous magnitudes of Larson’s Cube (LC), which contains the unit ball of the SUDR (r = 1/31/3) and is contained by the unit ball of the TUDR (r= 31/3), is equal to the total number of poles in the 3D tetraktys:

Equation 1:  (2^0 + 2^1 + 2^2 + 2^3) = 27

This is exciting because it relates the discrete numbers of the tetraktys to the continuous magnitudes of LC. But to understand it, one has to understand the equation above. It is an equation of what we call the Reciprocal System of Mathematics (RSM).

In the legacy system of mathematics, the sum of the numbers in equation 1 above is 15, not 27. In the RSM, the base number 2 in the binomial expansion is the magnitude’s number of “directions” (think polarities) and its exponent is the number of its dimensions.

This relates to Pascal’s triangle and Clifford algebras, but also to Raul Bott’s periodicity theorem, which limits the number of geometric dimensions to no more than three. The key is to understand that each number represents the number of geometric entities of 2n (n = 0-3), contained in a given dimension, where the dimension is the number of terms in the equation, less 1.

Thus, in equation 1 above, the total number of terms is 4, so the dimension is 3 and there is one 20 term contained in that dimension, three 21 terms, three 22 terms, and one 23 term. These happen to be the four coefficients of the fourth line of Pascal’s triangle, the sum of which is the mysterious number 8 of Bott periodicity:

Line 1) = 1        = 1
Line 2) = 11      = 2
Line 3) = 121    = 4
Line 4) = 1331  = 8

Since these numbers correspond to the first four dimensions of the binomial expansion and Clifford algebras, and they correspond to the geometry of Larson’s 2x2x2 stack of unit cubes, the philosophy of the ancient Greek tetraktys, and the dimensions of the only known normed division algebras, not to mention the blades of Hestenes’ Geometric Algebra, and the domain of string theory, our interpretation of them as the “directions” in the geometric dimensions of numeric magnitudes is in good company.

Hence, we can rewrite equation 1 above to reconcile the RSM and legacy math difference, by combining the coefficients of Pascal’s triangle with the terms of the binomial expansion:

Equation 2: 1(2^0) + 3(2^1) + 3(2^2) + 1(2^3) = 27

However, the coefficients of the fifth line of Pascal’s triangle, corresponding to the fourth dimension (5 terms less 1), are

Line 5) 14641 = 16

Obviously, the coefficients of each line of the triangle sum to powers of 2, while the sum of the products of the coefficients and the n dimensional geometric entities are powers of 3. For example , the sum of the products of the fourth dimension is:

Line 5) 1(2^0) + 4(2^1) + 6(2^2) + 4(2^3) + 1(2^4) = 3^4 = 81.

The coefficients of the triangle’s fifth dimension are

Line 6) 1 5 10 10 5 1 = 2^5 = 32

and the sum of the products of the coefficients and the n dimensional entities is

Line 6) 1(2^0) + 5(2^1) + 10(2^2) + 10(2^3) + 5(2^4) + 1(2^5)= 3^5 = 243.

But since our RSM interpretation counts the number of n-dimensional entities contained in each dimension, and those dimensions cannot exceed 3, what meaning can the triangle’s lines 5 and up have for the RSM?

Well it turns out to be the key to the RSM actually. Because the sum of the products of the triangle’s coefficients and the n-dimensional geometric entities is always a power of 3, we can simply multiply each coefficient of line four (the 3rd dimension) of the tetraktys by 3 to obtain the next higher set of coefficients, thus preserving the four terms of the three-dimensional form:

Line 5) 1331 x 3 = 3+9+9+3 = (8 x 3) = 24

Line 6) 3993 x 3 = 9+27+27+9 = (24 x 3) = 72

Line 7) 9 27 27 9 x 3 = 27+81+81+27 = (72 x 3) = 216

Notice that this is tantamount to multiplying the number of geometric entities of the LC by the number 3.

So much for obtaining the coefficients. Now, what about the sum of their products with the n-dimensional geometric entities?

Well, as we have already seen, these are just powers of 3:

Line 4) 3^3 = 27
Line 5) 3^4 = 81
Line 6) 3^5 = 243
Line 7) 3^6 = 729

So, we have another periodicity. This time it is a period of 27.  We will call it the tetraktys periodicity. Therefore, in the RSM, the number 8 of Bott periodicity, which is really the sum of the four coefficients of the triangle’s fourth line, is related to the number 27 of the tetraktys periodicity, which is really the sum of the four coefficients’ products with the corresponding four n-dimensional entities of the tetraktys (LC,) by the number 3, which is the key factor of both periodicities, uniting the discrete periodicity with the continuous periodicity.

This is a wonderful gift, a gift of 3, the trinity we might say, one which we neither understand nor deserve, but which we celebrate this time of year.

You lost me after the phrase "the next higher set of coefficients"
e.g. in this post you list Line 5 coefficienta several times, yet the coefficients are different in those cases. I'd expect, that the same line would always have the same coefficients.

Also, why are you fixated on cubes if Pascal triangle also works for Tetrahedron faces, edges, vertices, etc... ?

Do you know of "Pascal Simplex" ?

Do you know that Pascal Triangle can be extended to negative numbers?
Would they correspond to reciprocal entities?

December 17, 2012 | Horace

Horace wrote:

You lost me after the phrase "the next higher set of coefficients"
e.g. in this post you list Line 5 coefficienta several times, yet the coefficients are different in those cases. I'd expect, that the same line would always have the same coefficients.

What I'm saying is that to move up to the next higher dimension, just multiply a dimension's coefficients by 3. So, to calculate the coefficients of line 5, multiply those of line 4, by 3:

Line 4 x 3 = Line 5: 1331 x 3 = 3993 = 14641 = 81

See, 3993 are 3D coefficients, because they have four terms that are equivalent to Pascal's 4D coefficients that have five terms: 14641. When you multiply each of them out by their corresponding nD entities, they both sum to 3^4 = 81.

Horace wrote:

Also, why are you fixated on cubes if Pascal triangle also works for Tetrahedron faces, edges, vertices, etc... ?

Do you know of "Pascal Simplex" ?

Do you know that Pascal Triangle can be extended to negative numbers?
Would they correspond to reciprocal entities?

I don't know anything about those aspects of the triangle, but I am pursuing the relationship between discrete numbers and continuous magnitudes, which is at the heart of the trouble with physics.

If we write the reciprocal tetraktys as:

1/2^0
1/2^0 1/2^1
1/2^0 1/2^1 1/2^2
1/2^0 1/2^1 1/2^2 1/2^3

and then apply the 3D coefficients as calculated for line 5 (3993), then the ratio between the inverses is amazing (4^0, 4^1, 4^2, 4^3).

I'm still trying to make sure and then write an article on it.

December 17, 2012 | Doug

Line 4 x 3 = Line 5: 1331 x 3 = 3993 = 14641

See, 3993 are 3D coefficients, because they have four terms that are equivalent to Pascal's 4D coefficients that have five terms: 14641.

Doug!
You need to be less careless with these equations.
I might understand your concepts but others might not.
Writing such obvious untruths like 3993 = 14641 is a bad idea, that just decreases your credibility and makes it impossible to understand you by all except the closest insiders.

You must invent a new notation, that will make those equations understandable by all.

Now back to the conceptual realm:
Even I cannot understand how "3D coefficients "3993" are equivalent to 4D coefficients "14641" ".

Just because they can be manipulated by multiplication and both evaluate to 81...WTF?

December 19, 2012 | Horace

Yeah, you are right. I need to take more time to explain it repeatedly. I need to use a sign like "~" instead of "=" in those cases. I tried it once, but it didn't work any better.

I'm writing an article now to explain it more clearly, but, to be really helpful, I have to update the SPUD, which new-comers can use to go from A-Z and catch up. I have now started working on that.

You wrote:

Even I cannot understand how 3D coefficients "3993" are equivalent to 4D coefficients "14641"

I posted a new excellent video in LRC Lectures Online by John Baez that explains the LST view of the mathematics that is so important to understand, as background to what I am saying.

In his video, John talks about the tetraktys, but doesn't call it that. He tells the math history of imaginary numbers and their use to devise higher-dimensional algebras from the reals. He explains the important concept of normed division algebras and how they only exist in certain dimensions, and how that fact kept Hamilton frustrated for a long time, as he sought to find a three-dimensional normed division algebra (although he didn't have a label like that to express it concisely).

The problem is, Baez and company have changed the meaning of the word dimension, without explicitly pointing it out to non specialists. They don't necessarily refer to geometric dimensions, but like Larson explained, the word for them means the number of independent magnitudes in an equation.

Thus, complex numbers, Z, are two dimensional numbers for modern mathematicians, because they consist of a real and an imaginary part (a+ib), and they are used in 2D rotation.

However, in terms of the RSM, line 2 of the tetraktys (11), or 1(2^0), 1(2^1) in its binary expansion, defines a one-dimensional line:

(-)<----.---->(+)

2^0 = 1 is the magnitude of the point in the middle,1/1, and 2^1 = 2 is the one dimensional magnitude with two opposite polarities, or a dipole, stretching out on either side of it. In other words, it is the numerical equivalent of a geometric line.

Hamilton finally realized that he needed four terms to construct what he thought of as a 3D algebra, moving one dimension up from the 2D complex numbers. He realized he could do what he wanted to do with 1 real number and three imaginary numbers, or what mathematicians would call a 4D number.

Today, these numbers are called quaternions, H, and Baez refers briefly to the battle that ensued as to whether these "3D" quaternion numbers, or the "2D" complex numbers were the numbers of choice to use in science and engineering.

The champions of H lost, and the H's descended into the dusty bin of academics for a long time, but the important thing to understand is that they are not 3D numbers!

They correspond to the third line of the tetraktys (121), which, in binomial expansion form, is the numerical equivalent of a geometric 2D area: 1(2^0), 2(2^1), 1(2^2), because (I can't draw this in text) it consists of a point, and two, orthogonal, dipoles, forming one quadrupole.

Increasing line 3 of the tetraktys to line four (1331) gives us a true 3D numerical equivalent of a geometric volume, which, as Hamilton's friend Graves pointed out to him, could be formulated with seven imaginary numbers, but he didn't realize that these constructions, now called octonions, O, were in the true 3D realm of the tetraktys, which we at the LRC now know is the numeric equivalent of the 2x2x2 stack of unit cubes, we call Larson's Cube, containing one 2^0 point in the middle, three orthogonal dipoles, three orthogonal quadrupoles and one octopole, or 1(2^0), 3(2^1), 3(2^2), 1(2^3), in terms of the binomial expansion.

This is where the professor stomps his foot, because Baez and company apparently don't recognize this. Instead they conceive of the 8 corners of a single unit cube, not a 2x2x2 = 8 stack of unit cubes, and they assign 7 of the 8 corners of the cube (poles) to the 7 imaginary numbers of Graves, and from this they devise a way to concoct the last remaining normed division algebra.

It's complicated and a total distortion (they collapse the cube into a Fano plane) of the true nature of the tetraktys, as understood in terms of the binary expansion and its numerical equivalents of the geometric point, line, area and volume.

But what is very useful is the recognition that the sum of the numbers of the tetraktys, which defines Pascal's triangle,

1 = 1
1+1 = 2
1+2+1 = 4
1+3+3+1 = 8

and which they call dimensions, limits the number of normed division algebras to these four. This is the tetraktys and it contains all the geometric entities of our 3D universe, corresponding to point, line, area and volume. It's a case of numbers confirming the first postulate of the RST.

Moreover, because of their mathematical concept of the single cube, with its 8 corners, defining the algebra of octonions, they understand that the next line of Pascal's triangle (14641), leads to a fourth geometric dimension in the form of the first hypercube, and they don't stop there, but continue to higher dimensional hypercubes, collapsing them into a plane and studying their properties with group theory and Lie algebras.

To us this is a grievous mistake, not only because it misses the point of the identification of the binomial expansion with the magnitudes, dimensions and polarities of the tetraktys, geometrically expressed in the form of Larson's Cube, but also because it hides the real implications of Bott periodicity.

Baez explains this as the reason the number 8 is one of his three favorite numbers, but he can't explain why this number is so significant, no one can, unless they admit that the physical universe is limited to three dimensions, geometrically speaking, as stated in the RST fundamental postulates.

If they would only admit this limitation, they would understand that lines 5 and up of Pascal's triangle should be interpreted as compounds of line 4 of the triangle. In other words, lines 5 and up are multiples of line 4.

1) 1 = 1
2) 1+1 = 2
3) 1+2+1 = 4
4) 1+3+3+1 = 8
5) 2(1+3+3+1) ~ 1+4+6+4+1 = 16
6) 4(1+3+3+1) ~ 1+5+10+10+5+1 = 32
.
.
.

Thus, the true nature of Bott periodicity is seen in the scalar expansion of three dimensions in the form of Larson's Cube, which can best be understood as the linear expansion of a 1-unit dipole as the independent variable, with the 4 quadrupoles and 8 octopoles, as dependent variables.

Also, since two of the three dipoles and quadrupoles are degenerate in the 2x2x2 stack, we only need scale 1 each of them, as we go up each step of the triangle. Hence, starting with line 4 and renumbering:

1) 1^1(2^1) 1^2(2^2) 1^3(2^3) = 1 2x2x2 = 8, 1-unit cubes,
2) 2^1(2^1) 2^2(2^2) 2^3(2^3) = 1 4x4x4 = 64, 1-unit cubes,
3) 3^1(2^1) 3^2(2^2) 3^3(2^3) = 1 6x6x6 = 216, 1-unit cubes,
.
.
.

We can see from this that the doubling of the sums of the steps in Pascal's triangle, is due to the counting of the redundancy in the expansion of the 2x2x2 stack: there are ALWAYS 3 dipoles and 3 quadrupoles in the expanded stacks, so if we count all these entities, we get (counting just the number of entities, the bolded coefficients):

1) 1(2*0) + 3(2^1) + 3(2^2) + 1(2^3) = 8
2) 2(2*0) + 6(2^1) + 6(2^2) + 2(2^3) = 16
3) 4(2*0) + 12(2^1) + 12(2^2) + 4(2^3) = 32
.
.
.
Thus, we see clearly what has happened. The numbers in the expansion of Pascal's triangle is just a 3D expansion of Larson's cube, not an n-dimensional expansion of a hypercube, as Baez and company see it.

Neat, huh?

December 20, 2012 | Doug

This is an interesting quote from PhilG (Phillip Gibbs?)

Any mathematicians (sic) who has been around awhile knows of many surprising connections that have been found between parts of mathematics that at first seemed separate. Examples include modular forms, elliptic curves, exceptional groups, lattices, coding theory, special functions, cohomology, octonions and even quantum physics and superstrings. Many of these beautiful subjects have deep connections that suggest the existence of some unknown master structure that encompasses them all in some elegant way.

Of course, you know where I place my bet for what that "unknown master structure" is.

December 21, 2012 | Doug