Well, it is fun playing with circulants, but as Carl points out we might want to worry about idempotency also. For 1-circulants of the form

XYY

YXY

YYX

the only determinant 1 idempotent is the identity matrix. Proof: since $Y = Y^2 + 2XY$ then $X^2 = – 2Y^2 – \frac{1}{2} Y + \frac{1}{2}$, and from the relation $Y = Y^2 + 2XY$ it follows that $X = – \frac{1}{2} Y + \frac{1}{2}$. Thus $- 2 Y^2 = 2 Y^2$ and so $Y = 0$. By symmetry, the circulant matrices given by $(X,Y,Z) = (0,1,0)$ and $(0,0,1)$ also satisfy both the determinant and idempotency constraints. If we assume that $Y = \overline{Z} = r e^{i \theta}$ and that $X$ is real, it follows that the discriminant

$1 – 8X (a – ib)$

must be a positive real so that $b = 0$ and $Y = Z$ is real, and thus once again the identity is the only solution.

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## CarlBrannen said,

September 29, 2007 @ 11:09 pm

In general, there are exactly 8 idempotent 3×3 circulant matrices over the complex numbers. They’re discussed on pages 39-40 of my magnificient (but incomplete and now kind of dated) book, and the general solution is given as equation 3.39.

Of these 8, three are primitive. The other five are zero, unity, and the three matrices you get by taking the three primitives and adding them together two at a time.

The three primitives are a complete set, which means they sum to unity. Among the 3×3 matrices, one can find many such “complete sets of primitive idempotents”, and they all have this same structure of 8, which is given as figure 3.2 on page 37.

I’m fairly sure that all the complete sets of primitive idempotents of 3×3 matrices are equivalent under a transformation of the form P goes to S P S^{-1}.

If this is true, (and I’m wide awake right now so my error rate is somewhat reduced), then you can derive the structure of the idempotent circulants by looking at the 3×3 diagonal idempotent matrices (which have 0s and 1s on the diagonal and are elsewhere zero).

What’s confusing about all this is that the Koide mass matrices are not idempotent. The whole idea associating particles with primitive idempotents would suggest that there is no reason for the Koide matrices to be idempotent. Where they end up circulant comes from their arising from a color type force which is symmetric under even permutations (of the Feynman diagrams for their bound states).

The three primitive idempotent circulant matrices are what one ends up associating with the generation structure.

A mathematical detail I’m not certain about is this: Is it true that you can transform any complete set of primitive 3×3 idempotent matrices into another set (say the diagonal ones) by a e^{+S} P e^{-S} transformation? I suspect it’s true but haven’t worked on proving it. The exponential form is particularly useful because it allows one to control what parts of the Clifford algebra get transformed and which parts are left alone.

## Anonymous said,

September 30, 2007 @ 1:45 am

Thanks for the handy reference to your book. I’m still keen to play with general circulants a bit more, since the mass matrices are of this type. Maybe a knowledgable algebraist could give us a tip on the proof you’re after … or maybe I’ll get around to it sometime.

Remote Kea## a quantum diaries survivor said,

September 30, 2007 @ 8:39 am

Sorry for the off-topic but Kea, are you alright ? I heard there’s been a very strong quake down there. Let us know!

T.

## Kea said,

September 30, 2007 @ 9:11 pm

Well, it was a 7.4 but off the coast: hardly made the front page today. Here’s the story.

## Kea said,

September 30, 2007 @ 9:16 pm

Of course, if an earthquake that big happens off the east coast, I might have to start running damned fast.

## kneemo said,

October 1, 2007 @ 7:55 pm

Is it true that you can transform any complete set of primitive 3×3 idempotent matrices into another set (say the diagonal ones) by a e^{+S} P e^{-S} transformation?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.

## CarlBrannen said,

October 1, 2007 @ 9:51 pm

Nice comment, Kea, it really shows how I think SU(3) comes up from the color bound state.

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.

## Kea said,

October 1, 2007 @ 9:53 pm

That was kneemo!

## CarlBrannen said,

October 2, 2007 @ 1:32 am

Sorry, it is Michael who made the SU(3) comment. This is another publishable paper.

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.

## kneemo said,

October 2, 2007 @ 3:51 am

… SU(3) comes up from the color bound state.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.

## CarlBrannen said,

October 2, 2007 @ 10:11 am

Kneemo, maybe that sort of thing (using a SU(2) subgroup of SU(3) ), might be useful when you want to model a bound state that has two identical objects and one object that is distinct.

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.

## kneemo said,

October 2, 2007 @ 7:33 pm

Carl,

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.

## test said,

April 27, 2014 @ 2:25 am

Hmm is anyone else encountering problems with the pictures on this blog loading?

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.