# Precolor and Primitive Idempotents

An idempotent is an element of an algebra that is unchanged when it is squared, $\rho^2 = \rho$. In an algebra, the “primitive” idempotents are those that cannot be written as sums of nonzero idempotents. In a sense, these are like primes. In the density matrix formulation of quantum mechanics, the particle states are primitive idempotents. The part of quantum mechanics that is most tightly connected to spacetime geometry is the spin structure of the spin-1/2 fermions. The algebra that is used for this is the Dirac algebra. The Dirac algebra has four gamma matrices, $\gamma^0,\gamma^1,\gamma^2,\gamma^3$, respectively one for the four spacetime dimensions, t, x,y, and z. We will not be dealing with coordinates, so when we write “x”, we do not mean the coordinate value, but instead the related gamma matrix. The Dirac algebra is the same as the Clifford algebra C(3,1), with the addition that differential $\partial_x$ is associated with the Dirac gamma matrix x, and the similarly with the other 3 dimensions.

This isn’t enough dimensions for us. We need one more, we will call it s. This means the algebra we will be working in is generated by 5 “gamma matrices” instead of just 4. And we will rely on just the algebra, we will not bother ourselves with writing down an arbitrary representation in matrices.

Our algebra elements x,y,z,s,t satisfy the following rules:
(1) they square as follows: xx = yy = zz = ss = -tt = 1.
(2) they anticommute. That is, xy = -yx, xz = -zx, etc.
(3) You can multiply and add them together.
(4) We use complex multiples of sums of products.

An example element of the algebra is (1+zst)(1-ixy)/4. The scalar part of an element is the (real or complex) part of the number that isn’t multiplied by any of the matrix things. For the element (1+zst)(1-ixy)/4, the scalar part is 1/4. By the way, this element is idempotent, but not primitive idempotent.

Other than that, the algebra acts like matrix multiplication and addition: Multiplication distributes over addition, addition commutes, sometimes multiplicative inverses exist sometimes they don’t, additive inverses always exist. Multiplication and addition are associative.

So this is the algebra that I work with. The above explains one reason why I think it is important: it is the simplest generalization to the most important algebra of QED, the Dirac algebra. I have spent a good amount of time exploring the primitive idempotent structure of this algebra.

Interpreting the primitive idempotents as quantum states, their natural quantum numbers turn out to have hypercubic form, where the dimension of the hypercube depends on the size of the Clifford algebra. When you graph the weak hypercharge and weak isospin quantum numbers for the standard model fermions, the result becomes a cube if you assume that the quarks are midway states, mixtures of the components that make up the fermions. This gives the figure shown above, but what’s interesting is that the weak hypercharge and weak isospin quantum numbers, as well as electric charge and weak charge. If “weak charge” is unfamiliar to you, see Quarks Leptons and Gauge Fields by Kerson Huang.

It is not my intention here to imitate the use of the Dirac algebra but to add another dimension. Each time physics dives deeper into particles the rules change. What I want is a way of deriving the structure of the elementary particles from geometry alone. To do this, I have to define a force between the particles, from geometry alone.

From comparing the structure of the primitive idempotents of a Clifford algebra to the structure of the fermions of the Standard Model, I’ve found a way of organizing primitive idempotents of this C(4,1) Clifford Algebra to give the fermions. From there, I made a number of guesses on how to define a force between primitive idempotents.

The method I came to was to assign energy values to the components of the primitive idempotents. Here’s a sample primitive idempotent:
(1-zt)(1+ixy)(1+s)/8 = (1 – s – zt +ixy +zts -ixys -ixyzt -ixyzts)/8.
My method is to suppose that these 8 components correspond to contributions to a potential energy. The potential energy is defined by the sum of the squared magnitudes of the components, with each letter taking a value of 3 (that is not squared). For the above primitive idempotent, the associated potential energy is calculated as follows:

V = $|(1 - s - zt +ixy +zts -ixys -ixyzt -ixyzts)/8|^2$
= $(1/8)^2(1 + 3 + 9 + 9 + 27 + 27 + 81 + 243) = 400 / 64 = 25/4.$

The factor of 3 is a somewhat arbitrary guess. The actual values are hidden from us because the scale is the Planck mass. The observed particles are to be combinations of primitive idempotents that happen to cancel all their components leaving an approximate energy of zero. This is the model’s explanation for why the observed particles have masses so small compared to the Planck mass. In this model, the 8 components can be thought of as 8 different forces, with “1” being the weakest and “ixyzst” being the strongest. From a practical point of view, this (along with the factor of 3 for each reflection), defines the hierarchy of how primitive idempotents assemble themselves into the standard model fermions.

The 8 components also correspond to quantum numbers. For the primitive idempotents, there are only two possible quantum numbers, +1 and -1. Quantum numbers are additive, so if we have a particle made up of some combination of primitive idempotents, we can calculate its total quantum numbers by addition.

The equation of motion for space time is the 5-dimensional analog to the (massless) Dirac equation: $\gamma^\mu d_\mu \psi = 0$. This equation defines a relationship between the Clifford algebra and the spacetime coordinates. Let’s write the algebra elements with “hats” to distinguish them from the coordinates (because we briefly need to mention coordinates), and write the 5-dimensional Dirac equation, with the speed of light included, as follows:

(1) $\;\;\;(c(\hat{x}\partial_x + \hat{y}\partial_y + \hat{z}\partial_t + \hat{s}\partial_s) + \hat{t}\partial_t)\psi = 0$.

Suppose we are interested in a plane wave that is moving at speed u in the +z direction. Such a wave could be written as $\psi(x,y,z,s,t) = \psi(z-ut)$. Putting this into the 5-dimensional (or usual) Dirac equation, we find that such a wave can exist if u = +c or -c, and that the Dirac equation will be satisfied if the wave has a Clifford algebra value of $(1-\hat{z}\hat{t})/2\;\; A$, where “A” is any element of the algebra. We write the general solution in this form because $(1-\hat{z}\hat{t})/2$ is idempotent. Another way of saying this is that the plane waves in a given direction form an ideal of the algebra generated by (and we now revert to our previous hatless notation) the idempotent (1-zt)/2.

Since the real part of (1-zt)/2 is 1/2, there are (1/2)/(1/8) = 4 primitive idempotents that can travel in the +z direction. By “4”, what we really mean is that there are 4 independent degrees of freedom because there are different ways we can break (1-zt)/2 into four primitive idempotents. These alternative ways of breaking an idempotent into its primitive parts are just the same as the ways one can define the Dirac electron as a combination of four parts: electron/positron and spin up/down, or instead electron / positron and right/left handed; either way, we have 2×2 = 4 orthogonal states. In addition, we could look at the primitive idempotent structure of (1-xt)/2 and (1-yt)/2. These would also have four components.

The four primitive idempotents in the (1-zt)/2 idempotent are the fundamental particles. We will write them, leaving off the factor of 1/8, as:
A $= (1-zt)(1-s)(1-ixy) = 1-s-zt-ixy+zts+ixys+ixyzt-ixyzts$
B $= (1-zt)(1-s)(1-ixy) = 1-s-zt+ixy+zts-ixys-ixyzt+ixyzts$
C $= (1-zt)(1-s)(1-ixy) = 1+s-zt-ixy-zts-ixys+ixyzt+ixyzts$
D $= (1-zt)(1-s)(1-ixy) = 1+s-zt+ixy-zts+ixys-ixyzt-ixyzts$
We can cancel out the ixyzts force four ways:
A+C $= 2-2zt-2ixy+2ixyzt$
B+D $= 2-2zt+2ixy-2ixyzt$
A+B $= 2-2s-2zt+2zts$
C+D $= 2+2s-2zt-2zts$

We call these four particles “snuarks”. The snuarks come as +/- pairs in two types. The two pairs differ according to their highest potential energy component. The first two snuarks, A+C and B+D have a nonzero “ixyzt” component and are the higher energy pair. These snuarks will be the building blocks for the left and right handed electrons (and muons and taus). The other pair of snuarks, A+D and B+C have a highest potential energy component of “zts” which is smaller than “ixyzt.” This pair of snuarks will be the building blocks for the left and right handed neutrinos. The quarks are to be made from mixtures of snuarks.

The strongest remaining quantum number in the snuarks is “ixyzt.” The first two snuarks have quantum numbers of +2 and -2 for this component, and the last two have quantum numbers of zero. This structure is similar to the doublet plus double singlet structure of the SU(2) representations observed in the standard model. The snuarks are analogs to the ribbons of the Bilson-Thompson helon (braid) model. The ribbons comes in three types with charges of +1, 0, -1. The snuarks are different only in that the 0 charge snuark is duplicated. The helon model allows the charge on ribbons to migrate along the ribbon. This is equivalent to our assumption that quantum numbers are additive.

The snuarks carry six different quantum numbers, both “ixyzts” and “ixys” have been eliminated by the pairing that cancelled “ixyzts.” The remaining six are:
“1” is trivial (always 2).
“zt” is trivial (always -2).
“s” and “ixyzt” do not change with (1-zt) and follow the +2,0,0,-2 pattern.
“zt” and “ixy” change with (1-zt) and follow the +2,0,0,-2 pattern. I.e. if we were looking at the (1-xt)/2 idempotents, these last two operators / quantum numbers would be different.

The assumption was made here that the potential energies of the components of the primitive idempotents all have approximate Planck mass, and therefore the observed particles have to be mixtures with zero values for these components. To do this, we assume that nature allows particles to change from one form to another. This, the fundamental transformation, has to exist in order to allow the left and right handed electrons to interact with each other. This is the mass interaction, which is the origin of the title of this blog, “Mass.”

We assume that the fundamental transformation interaction allows changes only in the scalar part, and that the changes are to the sign. This sort of interaction will preserve the potential energy but change the form of the particle. Also as with any wave theory, we allow sign changes to the primitive idempotents. An example of this mass interaction on a snuark (and bringing back the factor of 1/8 we’ve ignored) is:
M(A+C) $= M(2-2zt-2ixy+2ixyzt)/8 = (-2-2zt-2ixy+2ixyzt)/8 == -(2+2zt+2ixy-2ixyzt)/8$
In the above, (2-2zt-2ixy+2ixyzt)/8 is an idempotent (not primitive because it can be written as the sum of two nonzero idempotents, A and C). But the result of the mass interaction, (2+2zt+2ixy-2ixyzt)/8, is not. Therefore the mass interaction does not transform snuarks to snuarks, but instead transforms snuarks to linear combinations of snuarks.

In the figure at the top of this blog post, the reader can verify that this is the pattern of the weak hypercharge and weak isospin quantum numbers of the elementary fermions. The elementary quarks end up with the same weak isospin numbers, but have weak hypercharge numbers spread between various fermion values.

I think that’s pretty much enough for now. To get from here to mass formulas is a more difficult road that requires a lot more explaining, not all of which has been written down. For those wishing to study things at a lower, more comprehensive level, there is an incomplete book on the web.

Avoiding a lot of more difficult math, there is a clear analogy between the precolor force described here and the color force. This says that we should look for analogies between the generation structure of the fermions and the resonance structure of the baryons. Just such an analogy has been found, between the resonance structure of the $\Lambda_{3/2-}$, that is, the lambda D03 states, $\Lambda(1520), \Lambda(1690), \Lambda(2325)$ and the charged fermions.

The Lambda is a baryon, composed of three different quarks, the u, d, and s. Perhaps the analogy is exact for this combination because when two or more of the quarks are identical, the analogy gets messy. The D03 resonances have J=3/2 which suggests that spin is identical between the snuarks. Finally, the “D” in the “D03” comes from the spherical harmonics we discussed in an earlier post here.

I had wanted to write this after going through the standard model from the ground up, but there seems to be some interest, and describing the Lie group / algebra structure of the standard model will probably take me another two posts, but with Kea been talking about it, I figured that typing the mess up here would be kinder than posting it on other people’s blogs or PF. Eventually I’ll get through the standard model.

Filed under heresy, physics, Uncategorized

### 6 responses to “Precolor and Primitive Idempotents”

1. Tony Smith

Carl, in your C(4,1) Clifford algebra model, I don’t understand where are the colors (red, blue, green) for color force SU(3).
Would you have to go to a larger Clifford algebra to get the colors included at the fundamental level ?

Also, you say:
“… The snuarks …[related to]… four primitive idempotents … come as +/- pairs in two types. …
The first two snuarks, A+C and B+D … are the higher energy pair. These snuarks will be the building blocks for the left and right handed electrons (and muons and taus).
The other pair of snuarks, A+D and B+C … will be the building blocks for the left and right handed neutrinos. …”.

Are your left and right handed neutrinos of equal mass ?

Tony Smith

2. dorigo

Hi Carl,

very nice post, I did not understand everything – and hey, my brain is still on vacation along with my body – but I look forward to having a closer look, and also will read your next ones.

Cheers,
T.

3. carlbrannen

Tony, your question about color is deep enough I should type up another post and try to explain it in detail.

Tommaso, I’m afraid to admit that I am considering playing chess competitively “again.” I say “again” because I was the top board on my high school chess team and we took the New Mexico state team chess championship in 1976-77. While I did play tournaments against people I thought were pretty good, I only played in one USCF rated tournament.

In the last 30 years, I’ve played maybe 3 games. The problem is that I’m very competitive (as a sport, I take chess almost as seriously as I would take a duel with samurai swords) and will have to practice for a few hours per day for a rather long time before I am willing to sit down with an opponent. This also means that I have to retire first.

4. Tony Smith

Carl, if you feel as you say about chess, consider this quote from the web:
“… There have been episodes of violence on Russian space missions, fist fights over chess games. When there was an ax murder over a chess game in a Russian polar station Russia banned chess in space. …”.
All this human passion over a game whose best players are machines – and which may in the not-so-distant future be solved (thus being shown to be in principle as trivial as tic-tac-toe), as was checkers earlier this year.

Tony Smith

5. Jay R. Yablon

Hi Tony,
Been trying to track you down for awhile. Please let me know how to get in touch, I am at jyablon@nycap.rr.com. Also, please check out my new blog at http://jayryablon.wordpress.com/
.
Best,
Jay.
PS: Carl and Tony, I also am a chess lover, though had to choose whether to devote my spare time to studying chess or physics. You can see which way I opted. If God gave me two lives, I’d do both.

6. Jay R. Yablon

Hi again Carl,

You diagram at the top jumped out at me. It is similar to a paper I had published in proceedings of the “Excited Baryons” Conference in 1988, see http://jayryablon.files.wordpress.com/2007/08/left-chiral-flavor-multiplet.pdf .

Jay.