I’m trying to determine if its a problem on my end or if it’s the blog.

Any feed-back would be greatly appreciated. ]]>

Let’s think of an SU(3) transformation as a permutation of the three primitive idempotents. Recall that any permutation is a product of transpositions. The SU(2) embeddings in SU(3) give rise to such transpositions, and we can compose them to recover a general SU(3) permutation, i.e., P’=S(MPM*)S*. Here is an explicit example, showing the action of the three SU(2) embeddings on the standard set of primitive idempotents, along with their corresponding transpositions.

]]>In the standard model, color is enough to make things not distinct, but in snuark theory, color is just a matter of orientation and is not preserved by the mass interaction.

The leptons being made from three identical snuarks, while the quarks are made from 1 or 2 charged snuarks and 2 or 1 neutral snuarks. And the same applies to the baryons, some of which are made from mixed quarks and others from identical quarks.

]]>I should also note that when you restrict the SU(3) transformations to one of its three SU(2) subgroups, a 3×3 SU(2) transformation leaves one primitive idempotent invariant, while transforming the other two.

This might be helpful when you want to transform only one of the particles of your bound state at a time. Combining two such SU(2) transformations from different embeddings will give you a general SU(3) transformation.

At the Lie algebra level, the three SU(2) embeddings arise from the number of ways to embed the Pauli matrices in the Gell-Mann matrices.

]]>I can’t believe how much stuff I am going to have to add to my book when I update it next.

You guys should be breaking stuff up into “minimum publishable units” and fattening up your CVs.

]]>I’ve been resting, I’m just about ready to make another blog post. Somehow that turn of phrase brings to mind a bowel movement.

]]>Ok, let’s set S=iH. Then a finite SU(3) transformation of a 3×3 complex matrix M can be written as:

M’=M+[S,M]+1/2![S,[S,M]]+…=e^{S}M e^{-S}.

A quick calculation shows that SU(3) transformations preserve idempotency:

(e^{S}P e^{-S})(e^{S}P e^{-S})=

e^{S}P^2 e^{-S}= e^{S}P e^{-S}

where I used unitarity and the idempotency of P.

Given a set of mutually annihilating idempotents P1,P2,P3, create a new set: e^{S}P1 e^{-S}, e^{S}P2 e^{-S}, e^{S}P3 e^{-S}. We can quickly show the new set is mutually annihilating by noting:

(e^{S}Pi e^{-S})(e^{S}Pj e^{-S})=

e^{S}PiPj e^{-S}=0.

Another useful, well-known fact is that SU(3) transformations preserve trace. So that given a primitive idempotent P, we have:

tr(e^{S}P e^{-S})=tr(P).

Now we can address your question about transforming a complete set of idempotents into the “diagonal” set of idempotents (i.e., the standard set).

In practice, we usually start with a 3×3 hermitian matrix A, and invoke the well-known theorem that A is unitarily equivalent to a diagonal matrix D. The diagonal matrix D is merely a matrix expanded in terms of the standard set of primitive idempotents with real coefficents. So the theorem is asserting that:

A = e^{S}(aP1+bP2+cP3)e^{-S}

for distinct eigenvalues a,b,c and standard primitive idempotent set P1,P2,P3. We can rewrite this as:

A = a e^{S}P1 e^{-S}+b e^{S}P2 e^{-S}+c e^{S}P3 e^{-S}.

This tells us that we can apply SU(3) transformations and transform the standard set to a new complete set of primitive idempotents, which gives the spectral decomposition of a general hermitian matrix with distinct eigenvalues.

So as long as your Koide matrices are hermitian with distinct eigenvalues, SU(3) transformations can always map the corresponding primitive idempotents back to the standard set.

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