What is new with Koide sum rules?

mitchell porter

Suppose we have six flavors of quark in an SU(4) gauge theory. For the moment, suppose there are no other quantum numbers... Then we will have a 6x6 yukawa matrix.

Next, suppose that these yukawas are flavon vevs, and that the flavon potential has a discrete symmetry generated by the four "sequential" (#73) and two "family" (#82) Brannen transformations, for particular values of δ.

And now, let us augment this "theory", so that the usual electric charges for the quarks arise or are introduced, and so that the usual Pati-Salam higgsing of SU(4) to SU(3) occurs. It seems that the first step should introduce a "checkerboard" texture to the 6x6 yukawa matrix, and then the second step should "double" the yukawa matrix, so there's one 6x6 yukawa checkerboard for three-color quarks and another 6x6 yukawa checkerboard for leptons.

Finally, let us suppose that the sequential symmetries dominate the quark yukawas, and the family symmetries dominate the lepton yukawas (though the residual family symmetries in the quark yukawas may be strong enough to produce recognizable Koide phases of 2/27 for up quarks and 4/27 for down quarks). This can give us the waterfall for the quarks, and the original Koide relation for the charged leptons.

edit: I think the first thing to do, would be to create the theory of the second paragraph. That would be practice at constructing a theory in which a Koide waterfall of masses arose from a 6x6 yukawa matrix.

arivero

Quote by mitchell porter

Finally, let us suppose that the sequential symmetries dominate the quark yukawas, and the family symmetries dominate the lepton yukawas (though the residual family symmetries in the quark yukawas may be strong enough to produce recognizable Koide phases of 2/27 for up quarks and 4/27 for down quarks). .

I ack that the publication of Żenczykowski paper has biased me to consider again the family symmetries. It could be that we are seeing different aspects of a larger discrete symmetry group... but how? The real problem is that it does not seem a permutation group because for any pair of masses we exchange, we find that one Koide equation keeps invariant but obviously others, containing only one of the two masses, are not preserved.

by the way, the publication of Phys. Rev. D 86, 117303 (2012) officially raises the number of cites of Brannen's and of myself on this topic!. I get a citation to hep-ph/0505220 so that the author can refer indirectly to internet forums with a "Brannen, as cited in...". And Carl gets a second citation, directly to http://brannenworks.com/MASSES2.pdf

mitchell porter

I think it would be instructive to express all the fermion pole masses as multiples of the Brannen mass parameter for the original Koide triple, M_L, and then use the relationship between the top mass and the Higgs VEV to express the latter in the same units. M_L is presumably the fundamental quantity in the waterfall (because it apparently comes from QCD or SQCD), but I don't think we've thought about how to get the Fermi scale from it. Yet surely this should be playing a role in our thinking about the Higgs.

mitchell porter

Today I was looking at two new and two old papers. The new papers are "Neutrino Mass and Mixing with Discrete Symmetry" by King and Luhn, and "Top-quark and neutrino composite Higgs bosons" by Adam Smetana. The old papers - well, one was a thesis, Francois Goffinet's thesis, "A bottom-up approach to fermion masses", and the other was the co-authored paper resulting from it, "A New Look at an Old Mass Relation".

Together, they should have something to say about how to extend the waterfall to the neutrinos, to the mixing angles, and to the properties of the Higgs sector. Goffinet's concept of "pseudo-mass" was invented precisely to link the Koide relation to mixing angles. King and Luhn review flavon models with discrete family symmetries, for the neutrino sector. And Smetana tries to get the Fermi scale by having both a top condensate and a neutrino condensate, in a broad class of models featuring a gauged flavor symmetry. To get the numbers right he ends up needing a large number of right-handed neutrinos, so probably he is still missing something essential, but it begins to make the connection I called for in the previous comment.

arivero

Quote by mitchell porter

Today I was looking at two new and two old papers. The new papers are "Neutrino Mass and Mixing with Discrete Symmetry" by King and Luhn,

[itex]S_4[/itex] is intriguing. It is equal to [itex]Z_2 \times Z_2 \ltimes S_3[/itex], and I wonder if these, say, four copies of [itex]S_3[/itex] could be the four copies acting in the waterfall. Also, where does it come from? Speculatively, could be a subgroup of SU(5)_flavour -the sBootstrap group-. And more speculatively, what about the permutations of the 4 components of a spinor?

arivero

Given that a lot of Koide stuff seems related (hat tip to de Vries and Brannen here) to this matrix

[tex]\begin{pmatrix}-\frac{2\,\mathrm{sin}\left( t\right) }{\sqrt{6}} & \frac{\mathrm{sin}\left( t\right) }{\sqrt{6}}+\frac{\mathrm{cos}\left( t\right) }{\sqrt{2}} & \frac{\mathrm{sin}\left( t\right) }{\sqrt{6}}-\frac{\mathrm{cos}\left( t\right) }{\sqrt{2}}\cr \frac{2\,\mathrm{cos}\left( t\right) }{\sqrt{6}} & \frac{\mathrm{sin}\left( t\right) }{\sqrt{2}}-\frac{\mathrm{cos}\left( t\right) }{\sqrt{6}} & -\frac{\mathrm{sin}\left( t\right) }{\sqrt{2}}-\frac{\mathrm{cos}\left( t\right) }{\sqrt{6}}\cr \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\end{pmatrix}[/tex]

I have setup a wxMaxima notebook to play with it. Not that I like Maxima, I used it in a VAX and it was already superseded by REDUCE when Mathematica come. But it comes with Ubuntu and has a graphical interface, which Reduce has not.

mitchell porter

Koide has a model of leptons with an S4 symmetry embedded in a flavor SU(3).

mitchell porter

Since this is a Koide thread we have to mention Zenczykowski's latest, though it is about "family triples", and not what I call the "sequential triples" of the waterfall... He's still building on the generalization of the e-mu-tau 2/9 parameter to u-c-t and d-s-b; he proposes that another parameter, which is just "1" for e-mu-tau, is also "1" for the quarks if you use Goffinet's concept of "pseudo-mass". If that's true it's a breakthrough, as well as a headache for the waterfall, because aren't we getting overloaded with too many relationships at once?

He mentions the usual problem, that these relations work best for low-energy masses. We've previously discussed Sumino's efforts to have family gauge bosons cancel out certain QED corrections, so that Koide's relation may be exact; but I was always curious about whether there might be some dual description of physics, in which, rather than thinking of the UV as fundamental, you thought of the physics as "IR + new degrees of freedom at a series of higher energies" - the idea being that the cause of Koide relations might be more transparent in this hypothetical "infrared first" formulation.

Well, I wonder if this paper by Davide Gaiotto (from January 2012) might be relevant: "Domain Walls for Two-Dimensional Renormalization Group Flows". "Renormalization Group domain walls are natural conformal interfaces between two CFTs related by an RG flow. The RG domain wall gives an exact relation between the operators in the UV and IR CFTs." It seems a tiny step towards what I had in mind.

arivero

Quote by mitchell porter

If that's true it's a breakthrough, as well as a headache for the waterfall,

The waterfall could happily miss the last triple, d-u-s, in exchange by one of the "standard" ones, but d-u-s does a better prediction of the down mass that d-s-b.

A motivation to follow this track could be, put all the quarks in the faces of a cube, such that all the equations of the waterfall are the faces that meet in some vertex. You will notice that his cube has a property, that opposite faces have opposite weak isospin. You can also notice that we only need three equations to fix the faces.

One of the vertexes of this cube is DSB, and of course is opposite vertex has the faces of the up-type quarks. This is the only axis that does not correspond to a waterfall symmetry, and on other hand the DUS vertex is the only axis which is used in both extremes.

Going to discrete groups, S4 is the group of permutations of the four "Z3 axis" in a cube, while the subgroup S3 is contained in four not-very-different ways, each of them being the permutations that keep one of such axis invariant (you can exchange fully the vertex by the opposite, to implement the Z2 subgroup of S3).

arivero

The real problem of S4 is to know the physics content, the objects we are permuting. The suggestion of putting quarks (or leptons) in the faces of a cube is rarely seen in the literature.

About PZ and pseudomasses, I think it is not very different of the initial objections to Harari-Haut-Weyers, they also do a similar trick, or a trick that can be interpreted as taking only the diagonal of the undiagonalised mass matrix.

By the way, I note that the abbreviations 2/27 and 4/27 are first used by Sheppeard in her note 342

mitchell porter

The pseudomass adds up the contributions of all the mass eigenstates to one of the weak eigenstates. So it looks like "sequential" triples, like in the waterfall, apply to mass eigenstates, family triples (as in the original Koide formula) apply to weak eigenstates, and this wasn't noticed until recently because, for charged leptons, the weak eigenstates are the same as the mass eigenstates.

For the quarks, we can then think of the waterfall as the dominant chain of relationships, and then the mixing parameters encode the rotation away from waterfall mass values, required to produce family triples with 2n/27 phases.

For the leptons, perhaps the family triples dominate, and a waterfall is weak or nonexistent. (I'm still not clear on whether right-handed neutrinos could have masses of the order of the quarks, as in #81, and then give rise to the observed small masses via seesaw.)

p.s. Chris Quigg had a paper yesterday - "Beyond Confinement" - in which 2/27 shows up as the exponent in a relation between the top mass and the nucleon mass, in a unified theory!

arivero

I hated macsyma and I am transferring my hate to its free twin, maxima, but still this is interesting. I got again the waterfall while I was trying to solve for the full S4 symmetric set of Koide equations.

This is, I was trying to find mass values in the six faces of a cube such of for each vertex we have a Koide equation. The group of rotations of the cube is S4; imposing Koide explictly breaks the symmetry as it gives different values to different faces. Not a very convincing motivation, but ok to play a little bit.

Now, in maxima. You define
K(x,y,z):=x^2+y^2+z^2-4*(x*y+y*z+z*x);
so that
expand(K(x,y,z)*K(x,-y,z)*K(x,y,-z)*K(x,-y,-z));
is a degree 8 even polynomial on three variables;
we put all of it

Code:

Q(x,y,z):=z^8-28*y^2*z^6-28*x^2*z^6+198*y^4*z^4-1172*x^2*y^2*z^4+198*x^4*z^4-28*y^6*z^2-1172*x^2*y^4*z^2-1172*x^4*y^2*
z^2-28*x^6*z^2+y^8-28*x^2*y^6+198*x^4*y^4-28*x^6*y^2+x^8;

and now we can use maxima "eliminate" to do the equivalent, I guess, of Sylvester matrix.
factor(eliminate([Q(1,x1,x2),Q(1,x2,x3)],[x2])) shows two terms that just validate x^2=x3^2. So in my first step I also canceled these factors:

Code:

step1:factor(eliminate([Q(1,x1,x2),Q(1,x2,x3)],[x2]))/(x3-x1)^8/(x3+x1)^8;

In the next step, factor(eliminate([paso1[1],Q(1,x3,x4)],[x3])) happens to have as a factor the polynomial for Q(1,x1,x4) and then it trivially tell us that the whole system of vertexes (1,x1,x2),(1,x2,x3),(1,x3,x4),(1,x4,x1) has solutions. As we want to exhibe actually some solution, I cancel the factor first

Code:

step2:factor(eliminate([step1[1],Q(1,x3,x4)],[x3]))/factor(Q(1,x1,x4)^12);

and now I cross each of the factors against the extant equation Q(1,x1,x4)

Code:

for i:1 thru 16 do (
pol:part(part(step2[1],i),1),
sl:solve([pol,Q(1,x1,x4)],[x1,x4]),
for k:1 thru length(sl) do (
  s:ev([x1,x4],sl[k]),
  if featurep(s[1],real) and featurep(s[2],real) then 
     (s:abs(s),
     if s[1]>s[2] then s:[s[2],s[1]],
     print(s,float(s)), 
     )
  )
);

The process solves to:
four complete x1,x2,x3,x4,x1 cycles
two sequences x1,x2,x3,x4
two sequences x1,x2,x3

The solutions for the waterfall triplets cbt and bcs appear in the list, numerically as sqrt(t)=10.12, sqrt(b)=1.464, sqrt(s)=0.267, with sqrt(c)=1, in one of the not-closing sequences. I guess I need an expert on discrete groups in order to understand what is going on.

PS. wow, now I notice that Q surely is the trick that Goffinet uses to avoid the square roots somewhere in his thesis.

mitchell porter

Exciting progress!

Quote by arivero

wow, now I notice that Q surely is the trick that Goffinet uses to avoid the square roots somewhere in his thesis.

Page 70 (section 3.2.1).

I guess I need an expert on discrete groups in order to understand what is going on.

A further step would be to look for an S4-symmetric potential where these solutions are the local minima.

arivero

Page 70 (section 3.2.1).[/QUOTE]

Indeed it is the same polynomial.

Code:

z^8-28*y^2*z^6-28*x^2*z^6+198*y^4*z^4-1172*x^2*y^2*z^4+198*x^4*z^4-28*y^6*z^2-1172*x^2*y^4*z^2-1172*x^4*y^2*
z^2-28*x^6*z^2+y^8-28*x^2*y^6+198*x^4*y^4-28*x^6*y^2+x^8;

Code:

  4         3         3        2  2             2        2  2       3
(%o3) z  - 28 y z  - 28 x z  + 198 y  z  - 1172 x y z  + 198 x  z  - 28 y  z
           2           2           3      4         3        2  2       3
 - 1172 x y  z - 1172 x  y z - 28 x  z + y  - 28 x y  + 198 x  y  - 28 x  y
    4
 + x

The minor improvement is that here we are sure that it is an ·"if and only if" relationship; Goffinet, in the text, was worried that the squaring could be introducing spureous solutions. As we have shown that this poly decomposes exactly in the product of the four possible sign combinations of Koide equation, now we are in position to grant that every solution of Goffinet's matrix version (3.30) of the equation is really a Koide solution.

Let me copy here this equation 3.30, setting the determinant of M as a function of the traces in M and M^2:

[tex]|M| = {2 \over 3 * 32^2} {(7 (Tr M)^2 - 8 Tr M^2)^2 \over Tr M} [/tex]

arivero

The following Mathematica code will help to find the solutions to the S4 symmetric Koide system.

If this post is the only one you are going to read, remember that we are organising three generations in opposite faces of a cube, and each corner must agree with Koide equation. A way to solve this is to fix one face, say to unity, and then check to four corners of this face.

Please use the code line-by-line; it is listed here without EOF separators! Also, please verify that you select the non common factor in each step, the order could change between versions of Mathematica (this is done with version 9.0 in the free trial period)

Code:

K[u_, v_, t_] := u u + v v + t t - 4 (u v + v t + t u)
G[m1_, m2_, m3_] =FullSimplify[K[Sqrt[m1], Sqrt[m2], Sqrt[m3]] K[-Sqrt[m1], Sqrt[m2], Sqrt[m3]] K[ Sqrt[m1], -Sqrt[m2], Sqrt[m3]] K[Sqrt[m1], Sqrt[m2], -Sqrt[m3]]]
Expand[G[1, a, b]]
step1 = FactorList[Resultant[G[1, x, y2], G[1, y2, x2], y2]]
Resultant[step1[[2, 1]], G[1, x2, y], x2]
step2a = FactorList[Resultant[step1[[3, 1]], G[1, x2, y], x2]]
step2b = FactorList[Resultant[step1[[4, 1]], G[1, x2, y], x2]]
step2 = {step2a[[2, 1]], step2a[[4, 1]], step2b[[3, 1]], 
   step2b[[4, 1]]};
step2[[1]]
s1 = N[Solve[{step2[[1]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
s2 = N[Solve[{step2[[2]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
s3 = N[Solve[{step2[[3]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
s4 = N[Solve[{step2[[4]] == 0, G[1, y, x] == 0, x >= 0, y >= 0, 
    x >= y}, {x, y}], 8]
sol = Join[s1, s2, s3, s4];
{1/x*174.1, y/x*174.1} /. sol

You can be intrigued that the solutions are more detailed than a simultaneus Solve[] of the system of four equations G[1, x, y2]==0, G[1, y2, x2]==0, G[1, x2, y]==0, G[1, y, x]==0. I am intrigued too. It seems that some of the particular solutions found by the Resultant method are embedded inside a continuous spectrum of solutions, and then the Solve method avoids listing them twice. I wished to know more on the relationship between resultants and continuous solutions.

It is amusing that the first triplet of the list is Rodejohann-Zhang triplet,
{1, {x -> 102.50258, y -> 2.1435935}}. Scale it times 174.1/102.50 and you get 1.69849, 174.1, 3.64088

My own triplet appears later, as it is generated by the last polynomial... it is
1, {x -> 2.1435935, y -> 0.071796770}. Use the same scale factor than before, and you get 1.69849, 3.64088, 0.12195

Both triplets are of the kind that becomes hidden in the continuous under Solve.

I am not sure about why this resolvent method does not find solutions with a zero, for instance 1.69849, 0.12195, 0. They can be searched by starting from G[0, x, y2]==0, G[0, y2, x2]==0, G[0, x2, y]==0, G[0, y, x]==0

Code:

IN:  Solve[{G[0, x, y2] == 0, G[0, y2, x2] == 0, G[0, x2, y] == 0, 
  G[0, y, x] == 0, x == 0.12195, G[r, x, y2] == 0, G[r, y2, x2] == 0, 
  G[r, x2, y] == 0, G[r, y, x] == 0, y >= y2}, {x, y, y2, x2, r}]

OUT: 
{x -> 0.12195, y -> 0.00875562, y2 -> 0.00875562, x2 -> 0.000628625,   r -> 0},
{x -> 0.12195, y -> 0.00875562, y2 -> 0.00875562,  x2 -> 0.12195, r -> 0}, 
{x -> 0.12195, y -> 0.00875562,  y2 -> 0.00875562, x2 -> 0.12195, r -> 0.261411},
{x -> 0.12195, y -> 0.00875562, y2 -> 0.00875562, x2 -> 0.12195,  r -> 3.13693}, 
{x -> 0.12195, y -> 1.69854, y2 -> 0.00875562,  x2 -> 0.12195, r -> 0}, 
{x -> 0.12195, y -> 1.69854, y2 -> 1.69854,  x2 -> 0.12195, r -> 0}, 
{x -> 0.12195, y -> 1.69854, y2 -> 1.69854,  x2 -> 0.12195, r -> 3.64099}, 
{x -> 0.12195, y -> 1.69854,  y2 -> 1.69854, x2 -> 0.12195, r -> 43.6919}, 
{x -> 0.12195,  y -> 1.69854, y2 -> 1.69854, x2 -> 23.6577, r -> 0}}

EDIT: ok, a faster recipe could be

Code:

pols:factor(eliminate([G(1,x,a),G(1,a,y)],[a]))/(y-x)^4$
f1:ev(part(pols[1],1),[y=x])$
float(sol1:solve([f1,G(1,a,x)],[x,a]));

the output has the following positive solutions:
[x = 29.85640584694755, a = 0.12453316162267], 
[x = 29.85640584694755, a = 650.4292237442922], 
[x = 29.85640646055102, a = 199.4974226119286], 
[x = 29.85640646055102, a = 13.92820323027551]
[x = 2.143593539448983, a = 102.5025773880714], 
[x = 2.143593539448983, a = 0.071796769724491], 

and thus
(%i58) mc
;
(%o58)                              1.69854
(%i59) mc * 2.143593;
(%o59)                           3.64097845422
(%i60) mc *  0.07179;
(%o60)                           0.1219381866
(%i61) mc * 102.50257;
(%o61)                          174.1047152478

mitchell porter

Quote by arivero

[tex]|M| = {2 \over 3 * 32^2} {(7 (Tr M)^2 - 8 Tr M^2)^2 \over Tr M} [/tex]

You could also: start with a 6x6 mass matrix including fictitious "up-down yukawas" as I have suggested, impose Goffinet's property on each of the four 3x3 blocks on the diagonal, and on the two larger blocks as in #82, and then finally impose a "checkerboard texture" in which all the "up-down yukawas" are set to zero, as in #73... and then see if the two larger blocks ever resemble the actual yukawa matrices.

Two problems: first, the SM yukawas are complex-valued and underdetermined by the experimental data (PDG). One would need to decide if the elements of the matrix M are the SM yukawas or secondary quantities derived from them. Second, the larger blocks are there in order to produce family Koide triplets, as in Zenczykowski; but Z's Koide triplets are made of Goffinet's pseudomasses, which are obtained by applying the CKM matrix to a vector of masses. It's not clear to me whether or not the larger blocks should be transformed somehow, before the Goffinet property is imposed.

arivero

Quote by arivero

Let me copy here this equation 3.30, setting the determinant of M as a function of the traces in M and M^2:

[tex]|M| = {2 \over 3 * 32^2} {(7 (Tr M)^2 - 8 Tr M^2)^2 \over Tr M} [/tex]

Just a thinking... Koide masses are fixed by an angle theta and a mass M_0, which is proportional to the trace. So if we suplement the above equation with the already know Tr M = 6 M_0, it takes a look very much as a the terms one usually sees in generalised Higgs potentials.

[tex]0 = {1 \over 1536} (252 m_0^2 - 8 Tr M^2)^2 - 6 m_0 Det(M)[/tex]

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