kneemo said:
Did you catch part where primitive idempotents came into the twistor string theory picture?
I saw it, can't claim to understand it. That issue with bilinearity reminds me of what Koide said about his formula. See the bottom of page 2 "Suggestion (A)" on this paper:
Challenge to the Mystery of the Charged Lepton Mass Formula
http://www.arxiv.org/abs/hep-ph/0506247
I should put a more philosophical explanation for why primitive idempotents should be studied with respect to the elementary particles:
The way that the standard model is put together is by making guesses at symmetries that apply to the elementary particles. If it weren't for relativity, the number of possible symmetries would be huge, but relativity cuts most of them down. But there is still quite a lot of freedom left.
One could look at the primitive idempotent structure of Clifford algebras as just another way of defining a symmetry but it's a little deeper than that. The primitive idempotents arise not from the symmetry of the elementary particles, but instead from the symmetry of a more primitive object, the equation of motion.
If you know the equations of motion, you can derive the symmetries but the reverse is not always true. In any case, the equations of motion are closer to the real world and if we are assuming that the real world is simple, then we should apply that rule to the equations of motion rather than to the symmetries.
As a particular example of this line of reasoning, the equation of motion of Newton's gravitation is exquisitely simple. The stuff that is conserved is not quite so. So since the equations of motion are simpler than the conservation laws, we may have more sucess finding the theory of everything by looking for simple equations of motion rather than looking for simple symmetries.
The basic idea here is that the equation of motion should provide very arbitrary motion, but that as humans, we call a "particle" a movement in the field that is conserved. We start with the usual Dirac equation which we will call the "spinor Dirac equation":
(\gamma^0\partial_t + \gamma^1\partial_x + \gamma^2\partial_y + \gamma^3\partial_z)\; \psi(x,y,z,t) = 0.
Previously we noted that moving from a spinor representation to a density matrix representation has the advantage of eliminating the unphysical U(1) gauge freedom.
In translating this change (spinor to density matrix) into the Dirac equation we have two choices. The usual method is to stick to pure density matrices, which results in a sort of double sided equation. The resulting equation doesn't increase the number of degrees of freedom beyond the usual density matrix form and so is boring as far as explaining symmetries between different particles.
The more obvious generalization of the Dirac equation to matrix form is seen less often. The idea is to simply replace the spinors with matrices. The resulting equation is the "matrix Dirac equation", which we will distinguish by capitalizing the wave function:
(\gamma^0\partial_t + \gamma^1\partial_x + \gamma^2\partial_y + \gamma^3\partial_z)\; \Psi(x,y,z,t) = 0.
The above two equations are identical except for the number of degrees of freedom held in the wave function. Instead of 4 complex degrees of freedom, the matrix has 16.
Suppose that we have a solution to matrix Dirac equation. There is an obvious way of extracting four spinor solutions from this one matrix solution, and that is to take each of the four columns of the matrix as a spinor.
A way of doing this splitting is to use the diagonal primitive idempotents, that is, the four matrices that are all zero except for a single 1 on the diagonal. For example, to extract the third spinor, we can use the diagonal projection operator with the 1 in the third position on the diagonal:
\Psi = \left(\begin{array}{cccc}<br />
\psi_{00}&\psi_{01}&\psi_{02}&\psi_{03}\\<br />
\psi_{10}&\psi_{11}&\psi_{12}&\psi_{13}\\<br />
\psi_{20}&\psi_{21}&\psi_{22}&\psi_{23}\\<br />
\psi_{30}&\psi_{31}&\psi_{32}&\psi_{33}\end{array}\right)
\psi_2 = \left(\begin{array}{c}<br />
\psi_{02}\\\psi_{12}\\\psi_{22}\\\psi_{32}\end{array}\right)<br />
= \Psi\;\rho_2 = \Psi\; \left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0\end{array}\right)
I've used density matrix "rho" notation because I want to point out how the spinors themselves are associated with the projection operators that pull them out of the matrix.
Turning the Dirac equation into a matrix equation gives us a very natural way of defining a multi-particle equation. From this we can derive geometric relationships between the particles. But we only have four spin-1/2 particles and this isn't enough for even one generation of fermions.
We can increase the number of particles held in the matrix by going to a more complicated Clifford algebra. This was done by Trayling, and later Baylis and Trayling:
A Geometric Approach to the Standard Model
http://www.arxiv.org/abs/hep-th/9912231
A geometric basis for the standard-model gauge group
http://www.arxiv.org/abs/hep-th/0103137
also see citations:
http://www.arxiv.org/cits/hep-th/0103137
But as you can see from reading the above, there is not a good explanation for why the particles are different. This question, why are the particles so different is the buggaboo at the heart of analyzing the symmetries of the Dirac equation as an explanation for the standard model. In addition, the above method provides no explanation for the number of generations.
Another problem with this method is that it requies a fairly large number of hidden dimensions (i.e. increases in the number of canonical basis vectors in the mathematical language, or increases in the number of spatial vectors \gamma^1, \gamma^2, \gamma^3 ... in the physics language) to get the spinor structure complicated enough to give the needed number of degrees of freedom. The reason for the large number of hidden dimension has to do with the structure of the primitive idempotents a topic I will turn to next.
In order to understand the structure of primitive idempotents (and therefore spinors) in a Clifford algebra, you have to read just a small amount of not very complicated mathematics. My source for this is:
Clifford Algebras and Spinors
London Mathematical Society Lecture Note Series 286
Pertti Lounesto, Cambridge University Press, 1997[?]
pp 226-228
https://www.amazon.com/gp/product/0521005515/?tag=pfamazon01-20
This theory is the basis for all those charts you've all seen that give what matrix groups the various real Clifford algebras are equivalent to. I try to keep close to the physics so rather than having my efforts depend on particular representations (or worse, use division algebras other than the complex numbers as elements of the matrices). You may be able to find a description of the structure of the primitive idempotents, by looking on the web for "Radon Hurwitz Clifford". Also, the primitive idempotents define the ideals of the Clifford algebra so you can look for that too. But the above book is sufficiently useful and easy to read that I think everyone should own a well thumbed copy.
To go further I need to define a "mutually annihilating set" of primitive idempotents. These are a set of primitive idempotents that (a) add up to unity, and (b) give zero when any two different ones are multiplied. The diagonal primitive idempotents are clearly a mutually annihilating set. But it turns out that given any primitive idempotent, one can always find a set of mutually annihilating primitive idempotents which it is a member of.
The Radon Hurwitz numbers tell us why Trayling and Baylis had to add so many hidden dimensions in order to model just the 1st generation of fermions. It (approximately) turns out that if you add two hidden dimensions, you end up doubling the number of primitive idempotents in a "mutually annihilating" set. The reason for the (more or less) doubling requiring two extra dimensions is because each time we add a hidden dimension we double the number of degrees of freedom in \Psi, but in order to double a spinor we have to double the matrix, and doubling the size of a matrix increases its number of degrees of freedom by four. Thus adding two dimensions makes the matrices big enough to double the size of the spinors.
The way I prefer to organize the project from the other side. Instead of taking the elementary particles and packing them into matrices as if they were commuters on the Tokyo subway, I think it is more elegant to look at what the natural symmetries between the different idempotents of the Dirac equation and to assume that these are preons that make up the elementary particles.
There is another reason for approaching the problem this way. If you spend some time playing around extending the Dirac algebra into a more general Clifford algebra by adding hidden dimensions, you will find that the canonical basis elements (i.e. "Dirac bilinears") that you get are very naturally classified into four different types according to their symmetry with respect to the 3 spatial coordinates: [x, y, z, 1].
The mapping is done by first assigning the canonical basis vectors to the above four categories, and then assigning products according to the very simple and obvious rule:
\begin{array}{ccccc}<br />
\times&1&x&y&z\\<br />
1&1&x&y&z\\<br />
x&x&1&z&y\\<br />
y&y&z&1&x\\<br />
z&z&y&x&1\end{array}
One can easily verify that the above multiplication table is consistent with Clifford algebra multilication so that it gives a consistent classification of the canonical basis elements (Dirac bilinears) and therefore a consistent classification of the degrees of freedom of Psi.
Given any particular Clifford algebra, a set of mutually annihilating primitive idempotents is defined by a set of what the mathematicians call "commuting roots of unity". I've been calling them "normal operators" but that's lousy notation. In either case, they are canonical basis elements that square to unity and commute. When considered as operators acting on the set of primitive idempotents (i.e. density matrices) that they generate by (either side) multilication, they produce eigenvalues of +/- 1. It is these operators that I assume to be the natural operators for the primitive idempotents. As a concrete example of a set of commuting roots of unity in the Dirac algebra, each of the following sets is such a set (I use (-+++) signature if you don't then multiply everything by i):
\begin{array}{rccl}<br />
(&i\gamma^0,& i\gamma^1\gamma^2&)\\<br />
(&\gamma^1,& \gamma^0\gamma^3&)\\<br />
(&i\gamma^0\gamma^1\gamma^2\gamma^3,& \gamma^0\gamma^1&)\\<br />
(&0.6\gamma^2+0.8\gamma^3,&\gamma^0\gamma^3&)\end{array}
There is a big problem here. As you can see from the above (which really only scratches the surface of the variation in available in commuting roots of unity) any given Clifford algebra has an infinite number of distinct sets of commuting roots of unity. Consequently, we don't have any reason to choose one over another.
In assuming that the elementary particles are made up of preons defined by primitive idempotents, we also have the problem that the set of commuting roots of unity that define one set of primitive idempotents may not be compatible (in the quantum mechanical sense of commuting operators) from the commuting roots of unity that define another preon in the same particle.
Given any two primitive idempotents, there may or may not be a set of commuting roots of unity that define them. It is easy to determine if the two primitive idempotents are compatible. If they multiply to zero (or are identical) then they are compatible, otherwise not.
Gosh, I think this is more than enough about primitive idempotents and Clifford algebras.
Carl
I knew I needed to get into those exceptional algebras for some reason!
A good way to approach this beast is by seeking advice from one of its greatest hunters, Ed Witten. Back in Dec. 2003 Witten showed http://arxiv.org/abs/hep-th/0312171" that the perturbative expansion of \mathcal{N}=4 super Yang-Mills theory is equivalent to the D-instanton expansion of the topological B model with target space \mathbb{CP}^{3|4}. (selfAdjoint referred me to this paper originally)
