# Koide formulas and Qubit / Qutrit MUBs

Recently we’ve been discussing Mutually Unbiased Bases or MUBs on this blog. This has not been for any particular interest in pure mathematics, but instead in their application to a preon theory of the elementary particles, and the E8 quantum numbers. The relationship is rather long and difficult to explain. All the major pieces are already on the web in various places, but it would be rather difficult for someone to piece together the details as they are spread around, and use various notation. The full explanation is too long for a blog post, but what I can do is give an overview of the explanation, along with links to resources. I will begin with the end, the Koide formula, as it hasn’t been discussed here in a few weeks:

Koide Formulas: The Koide formula for the charged leptons that we will use here is: $\sqrt{m_n} = \mu_1(1+\sqrt{2}\cos(2in\pi/3 + 2/9) )$
where $\mu_1$ is a constant with units of square root of mass, n is the generation number and $m_n$ is the mass of the generation n charged lepton, that is, the masses of the electron, muon, and tau. The formula is accurate to a part in a million.

There is a similar formula for the neutrinos (i.e. the neutral leptons): $\sqrt{m_n} = \mu_0(1+\sqrt{2}\cos(2in\pi/3 + \pi/12 + 2/9 ) ).$
The accuracy of the neutrino formula is unknown, it accounts for the differences in masses of the neutrinos, but those masses are rather poorly measured.

The above is a slight rewrite of Koide’s original formula of 1982 (for the charged leptons), which is exact to experimental error (quite a coup from a time when the tau mass was not well known): Various efforts have been made to give a theoretical explanation for the formula. I think Koide’s paper of 2005 gives a pretty good summation of the problem: hep-ph/0506247; it uses the square root of mass; it is symmetric under permutation of generation; and the formula is for the pole masses.

The bilinear nature of the Koide equation suggests it should arise from the bilinear density matrix formulation of quantum mechanics rather than the linear state vector formulation. If the theory were explained using the usual symmetry breaking techniques we would expect it to be good for the unrenormalized masses but instead it is good for the zero temperature measured masses. This suggests that symmetry breaking will not explain this formula. The formula is amazingly exact and this suggests that it should arise from non perturbational methods rather than a perturbation series. So the nature of the formula suggests non perturbational calculations using density matrices. In fact, that is how I came upon the cosine version of the formula and its generation to the neutrinos.

Clifford Particles and Quantum Numbers

The Pauli and Dirac algebras are examples of Clifford algebras. Both the Pauli and Dirac algebras are directly connected to the geometric structure of space-time. The Pauli matrices relate to the 3 spatial dimensions, while the Dirac gamma matrices relate to the 4 coordinates of spacetime. It is possible to work on Clifford algebras that are not connected to spacetime, and this is done in elementary particles when the Pauli algebra is used to model the internal symmetries of particles such as isospin.

When the Clifford algebra is related to spacetime, as with the Pauli or Dirac algebras for spin-1/2 non relativistic or relativistic particles, the resulting symmetry is called an “external symmetry,” while the ones that have no spacetime associated are called internal symmetries. Elementary particles has postulated internal symmetries that do not correspond to a Clifford algebra, but since Clifford algebras are equivalent to matrix groups, one can always suppose that these non Clifford internal symmetries arise as subgroups of a Clifford algebra.

Suppose spacetime has hidden spatial dimensions. It’s natural to expect that the hidden dimensions will increase the size of the “external symmetry” Clifford algebra, and that this could be a way of unifying the elementary particles. In attempting to fit the elementary particles into this sort of scheme, one must first understand the spinor structure of Clifford algebras.

Since elementary particles are described by quantum numbers, one seeks to understand the structure of the natural quantum numbers of Clifford algebras. A short description of the structure is that the external particles form hypercubes: With the Pauli algebra there are two particles with opposite spins. With the Dirac algebra there are four particles with two possible spins and two possible charges. With higher Clifford algebras the number of particles will be some power of 2 (look up Radon-Hurwitz numbers for details), say $2^n$ with n “natural” quantum numbers distinguishing them. Each particle will take two possible values of that quantum number, say $\pm 1$ or $\pm 1/2$.

A Preon Model for the Fermions

Naturally, one next examines the elementary particles to see if their quantum numbers come in hypercube form. This turns out to be the case. If one graphs the 1st generation quarks and leptons according to their weak hypercharge ( $t_0$) and weak isospin ( $t_3$) quantum numbers, sure enough one finds an obvious cube. From my 2004 paper on the subject, “The Geometry of Fermions”, which is somewhat dated but gives a correct analysis of the Clifford algebra details, the 1st generation fermions graph as follows: The quarks are located on the vertical lines of the cube, while the leptons are at the corners. Furthermore, the quarks are triplets. This makes the structure of each vertical line have a color structure of {1,3,3,1} which suggests that both the quarks and leptons are composed of three preons as follows: In the above, the quarks appear with three different colors according to their composition.

So the implication of this simple model of the quarks and leptons is that the leptons are composite. The Koide formula results from the interactions of these preons, but to derive the formula we have to deal with a lot more Clifford algebra. The calculation will have to be non perturbative, and bilinear, so we next discuss the applications of density matrices to non perturbative QFT.

Obscure Density Matrix Theory

It is possible in low energy quantum mechanics to treat the proton as a spin-1/2 fermion. As energies increase, this simple model breaks down and one must replace this model of the proton with one that is built from three spin-1/2 quarks plus sea particles gluing them together. But at low enough energies, spin-1/2 is sufficient to model the proton. Quantum mechanics applies equally well to the bound state and the states that are bound. To fully exploit this fact, we need to have a theory that allows us to move from free states of subparticles to bound states.

The usual way of making a first order model of a quantum mechanical bound state of three spin-1/2 preons would be to take the tensor product, with proper symmetrization. We already know that this sort of thing is suitable for perturbation theory but the bound states we wish to model are very deeply bound and perturbation theory won’t work for them. In addition, the Koide formula is non perturbational and bilinear. Rather than use spinors, we will therefore build the bound state from density matrices. To explain how this works, we have to first extend a few concepts in quantum field theory slightly beyond where the reader may be familiar with them.

Our first extension is an observation of Schwinger from the 1950s that density matrices can be thought of as products of creation and annihilation operators. His notation is somewhat different from the modern, and from that used here, but the interested reader is invited to read his early papers on the subject, The Algebra of Microscopic Measurements (1959), and The Geometry of Quantum States (1960). Where Schwinger writes M(b,b) we interpret as the pure density matrix for the state b. More generally, Schwinger’s measurement symbol M(b,c) we interpret as the product of the creation operator for the state b times the annihilation operator for the state c, i.e. $b^{\dag}c$. Note that these more general density matrices are not Hermitian. As Feynman diagrams, they correspond to the situation where a boson has modified a fermion from the state c to the state b, but where the boson is not kept track of.

Our second extension is that in terms of Feynman diagrams, density matrices correspond to the propagators of virtual states. This is probably going to be obvious to those who work in the foundations of QFT, for others, a review of the Feynman diagrams of qubit QFT may help. See the paper by Iwo Bialynicki-Birula and Tomasz Sowinski, Quantum Electrodynamics of Qubits, 0705.2121, in particular the equation for the electron propagator, (38e). For a longer explanation in the context of how we will be using density matrices in this post, see the previous post on the subject, Feynman Diagrams for the Masses Part 2.

With these convenient extensions of the usual QFT in our toolkit, we can approach the problem of modeling the propagator (density matrix) for a bound state as a matrix whose elements are composed of density matrices for the states which are being bound, plus density matrices for the interactions between the bound states (ignoring the details of what happens to the bosons that carry the interactions).

The Algebra of Feynman Diagrams

As is usual with QFT, we will be working in momentum space, the Fourier transform of position space. The Feynman diagrams we will work with are particularly simple, they are the density matrices. Each has only two legs, an initial state and a final state. If we assemble them into more complicated diagrams, those diagrams will either be circles, or will also have only two legs, an initial and a final. Here is a diagram of pretty much the only thing we can do with these simple propagators: In words, a red fermion has been transformed into a green one by a boson interaction. We will handle the math for the fermions only. It turns out that we will be able to wrap all the boson mathematics into a term to apply at the vertex as will be discussed below.

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Note on Notation: We will use “x”, “y”, and “z” for the Pauli spin matrices. This will reduce the hassle of typing up this post considerably. In my LaTeX papers, Clifford algebra elements are written with hats over their head, so the Pauli spin matrices would be $\hat{x}, \hat{y}, \hat{z}$.

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In computing the combined propagator (the red / green propagator on the right above) from the two separate propagators, we must integrate over all possible positions for the vertex. We are in qubit QFT so this reduces to an integral over time, there are no position coordinates. And we are dealing with the Fourier transform, so, using the usual rule that the Fourier transform of a convolution is a product, we find that the combined propagator is the product of the two individual ones.

If we equate “red” with x and “green” with y in the Pauli algebra (which amounts to taking our states from the Pauli MUBs rather than dividing the problem up (in spinor fashion) into spin up and down), then we can write the above three propagators as (1+x)/2 for the red, (1+y)/2 for the green, and the combination propagator is represented by the non Hermitian pure density matrix which is twice the product of these: (1+x+y-xy)/2. The product is consistent with the Feynman rules for qubit propagators. Thus multiplication of density matrices is equivalent to the action of gluing our Feynman diagrams together. As is usual, addition corresponds to alternative paths that connect the same pair of initial and final states.

As an example of a bound state to which we’d like to apply these tools, a baryon is composed of three quarks with different colors. However, the three quarks can emit / absorb gluons and consequently change color. With our simplistic Feynman diagrams, we need propagators / density matrices corresponding to red, green, and blue, plus the 6 possible changes between these, a total of nine propagators: Because we are working in qubit QFT momentum space, the rules for connecting these Feynman diagrams together is simply multiplication, however, we want to put all the physical content (as far as what happens when a quark or preon changes color) into the above 9 diagrams, so when we glue them together, we will be careful to match the colors of the initial and final states that we hook up: We will only connect a red initial state up to a red final state, etc.

It is a convenient fact that the three quarks inside a proton do not annihilate each other, but instead simply interact by exchanging gluons. So rather than use the above symbols for the very short distance (free or single interaction) propagators, we can also use them for the very long term quark propagators. This is an inherently non perturbative assumption and it makes our calculation inherently non perturbative. But rather than compute the non perturbative propagators up by integrating over an infinite number of perturbations, we’re simply going to assume that this has already been done, and deal with the long term propagators only. (By the way, qubit Feynman diagrams are always finite, to all order in perturbation theory, as the reader who clicks and reads the references will know.)

So our Feynman diagrams for the baryon propagator will consist of 3 propagators taken from one of the above nine. Furthermore, each color has to appear once as an initial state and once as a final state. Other than that, anything goes. For example, any Feynman diagram that happens to leave all three colors unchanged will be summed up into the three individual single color propagators: To keep track of our Feynman propagators and to make sure that we only hook red up to red, we can think of them as put into a 3×3 matrix where the rows and columns correspond to the three colors. Because of the restriction on colors, any given Feynman diagram will only contribute to three out of the nine matrix entries. For example, the above Feynman diagram would contribute to the diagonal values as it takes red to red, green to green, and blue to blue. The color changing diagrams contribute to the off diagonal entries.

The matrix form is particularly convenient in that when we multiply two such matrices, the Feynman diagrams will be properly connected together by the rules of matrix multiplication. The longer post describing this process is Bound States as Density Matrices.

The matrix form is also convenient for the purpose of representing the bound state as a density matrix (or Feynman propagtor) for the baryon. As usual, we require that the density matrix be pure, and therefore that it be idempotent and have trace 1. To the extent that we can model Feynman propagators as complex numbers, this tells us to model the bound state as a 3×3 matrix that is a primitive idempotent. The idempotency relation amounts to the fact that the long term bound state is stable, so its propagtor does not change over twice that length of time.

The problem of classifying the primitive idempotents of 3×3 complex matrices is easy and is given in my still incomplete book on density matrices towards the end of chapter 3. For the case of the leptons, where the three colors correspond to three preons identical except for color, the color symmetry means that the matrix must be cyclic. The cyclic 3×3 primitive idempotent matrices are: where w is any of the three cubed roots of unity, $\exp(2i m\pi/3)$ for m=1,2,3.

This result is convenient because it tells us that the results of our calculations will come in 3s, which we relate to the three generations. Applied to the baryons, it states that baryon resonances also come in 3s, and a generalized Koide relation applies to them. The subject is further discussed in a previous post, Regge Trajectories and Koide’s Formula.

Mass and Generations

The three cyclic primitive idempotent matrices all share the same diagonal values. In terms of Feynman diagrams that says that they act equally as far as red to red, green to green, and blue to blue. Where they differ is in the interactions that bind them together, and these interactions have three consistent available levels; the generations differ in the sea preons, not in the valence preons. Another way of putting it is that the generations differ in their boson content, their fermion content is identical. And their average boson content is zero so they differ primarily in mass only.

A field amplitude must be squared to give a probability and so energy is proportional to the square of an amplitude (as in the case of the electric field represented by photons). So to compute the masses of the three generations, we have to sum over the field amplitudes. The three colors will contribute equally, but orthogonally, so we may as well deal only with the top row of the matrix and triple the result. We get: The above is fairly similar to the Koide relation as observed for the leptons (at the top of this post), but it differs in two major respects. First, there is an extra factor of the square root of 2 in the ratio of the contribution of the diagonal elements to the off diagonal elements, and second there is a missing phase angle of 2/9. To explain where these come from, we have to go back and fix up the boson vertices on our Feynman diagrams. Also, there is a mass scale that needs to be included.

As an example off diagonal element of the matrix, let’s deal with the red / green entry, that is, the one that takes a green initial state and turns it into a red final state. Since the matrix form of the Feynman diagram gives correct multiplication for any sort of collection of these Feynman diagrams, it also is useful when we are considering the effect of a single boson interaction, the one that converts green to red.

To compute the correct vertex for this operation, we need to make assumptions about these preon states. The simplest assumption is that red, green, and blue are Pauli states taken from three mutually unbiased bases (MUBs). The reason the MUB assumption is convenient is that it automatically defines both the amplitudes and the phases of the boson interactions. Between two Pauli MUBs, the transition probability is 1/2 so the off diagonal amplitudes need to be adjusted by a factor of $\sqrt{1/2}$. This brings the formula a little closer to the Koide form: $1 + \sqrt{2}\cos(2in\pi/3)$.

That Damned Number

The extra phase angle of 2/9 is more difficult, and was referred in the correspondence of the people working on this, Kea and Kneemo, as “that damned number” or TDN. Last year, I attempted to derive TDN from the assumption that it was a result of multiple interactions and a Berry phase. This didn’t work. Instead, it seems to come from the inclusion of virtual infrared bosons in the color changing reaction.

Rather than consider color changes that are exactly from green to red, we need to also consider the possibility that the system will change from green to near red, and then emit a lower energy boson to make the remaining transition to red. Thus we will break the interaction into one hard boson and an unknown number of soft ones. Following the example of chapter 6.5 of Peskin & Schroeder, “Summation and Interpretation of Infrared Divergences”, we find (equation 6.80) that the sum over all the possible virtual contributions gives an exponential modification of the bare single value: I’m not sure how to modify this for the present case, but it seems possible that the factor $\exp(2i/9)$ comes from such an exponential. If so, by the rules of how one computes Berry phase for these complex 3×3 matrices, the value for X would have to be $2i/3$.

Berry Phase

The other source of phase in the Feynman diagrams comes from Berry, or quantum phase. Berry phase is the phase that arises in a quantum system when it is sent through a series of orientation changes. In a MUB, Berry phase is a characteristic of products that involve states taken from three different bases. For example, the product (1+x)/2 (1+y)/2 (1+z)/2 is equal to (1+x)/2 (1+z)/2 multiplied by a complex number, $\sqrt{0.5}\exp(\pi/4)$ if I recall. For spin-1/2, Berry phase can be easily calculated as half the solid angle subtended by the three spin orientations. For x, y, and z, the solid angle is one octant or $4 \pi/8 = \pi/2$, so the Berry phase is $\pi/4$.

So long as one leaves the Feynman propagators in density matrix form there is no reason to have to calculate Berry phase; it will be handled automatically. But if we were to leave the Feynman propagators in density matrix form, then instead of dealing with complex 3×3 matrices we’d be dealing with 3×3 matrices of Pauli operators and our cyclic primitive idempotent calculations would not be correct.

In converting the MUB density matrices into complex numbers we are rewriting a finite subalgebra of a non commutative algebra in the form of a commutative algebra. It is a little surprising that this can be done. The technique for doing this in the present case is given in my book on density matrices, in the later sections of chapter 7. One finds that one can absorb the $\exp(i\pi/4)$ Berry phase into the complex matrix if one multiplies the off diagonal terms by $\exp(\pm i\pi/12)$. This phase does not appear in the electron, but it does show up in the neutrino.

Electrons and Neutrino Differences

The presence of Berry phase in the formula for the neutrino masses while not in the formula for the charged leptons is related to the fact that the Fermion cube (shown towards the top of this post) is not straight on, but is angled by 45degrees. The easiest way of accomplishing this rotation is to suppose that the 8 preons that make up the corners of the cube (and that contribute to the leptons without mixing) are all composite objects made from two subpreons each.

In the electron case, the two subpreons have oppositely directed Berry phases and the combined particle carries a Berry phase of zero. In the neutrino case, the two contributions are the same and the composite preon gets a Berry phase.

As is discussed in the references above, when Baryon triplets are written in Koide matrix form, they also require the angle 2/9. However, their Berry phases are different. Instead of $\pm \pi/12$, they take values of $\pm 3n \pi/12 = \pm n\pi/4$, where n runs from -2 to +2. This difference is apparently due to the fact that the baryons are color neutral, and so Berry phase must be equivalent to 0 mod 3.

Preon Quantum Numbers

Garrett Lisi’s fitting of the elementary particles into an E8 structure is attractive at least partly because the interactions are given a particularly simple representation in terms of the quantum numbers. If the quantum numbers of three particles add to zero, then they interact. We can compare this scheme with the natural quantum numbers of the preons we’ve defined here.

Rather than think of the mixed propagators as fermions modified by bosons, we can instead think of them as representations of the boson that does the modification. For example, the red / green quark propagator could stand for a RG-bar gluon. This sort of interpretation is similar to Garrett Lisi’s E8 classification in that it would mix the fermions and bosons. In this case, the fermions are represented by the usual pure Hermitian density matrices, while the bosons are represented by pure non Hermitian density matrices.

The presence of non Hermiticity in the boson representation might be a way of getting a generation structure into Garrett’s model. With red designated as (1+x)/2 and green given by (1+y)/2, the non Hermitian pure density matrix for the RG-bar gluon given here is twice the product, (1+x+y+xy)/2. In this model, the presence of red and green as annihilators as opposed to creators is entirely defined by their presence at the beginning or end of the product. To get a consistent quantum numbers out of them in the spinor form that Garrett uses (where annihilators and creators are split and take opposite quantum numbers) we can use x and y as the quantum numbers, but we have to (somewhat arbitrarily) choose to negate one of them, and with the gluon, we will have to delete the inconvenient xy term in favor. In all cases we would ignore the “1” part of the density matrices. A better scheme would be for three particles to interact if they sum to one another, ignoring the scalar component 1, and the non Hermitian component ixy. It seems likely that Garrett’s scheme could be rewritten this way.

Generation Quantum Numbers

Along the same line, we could write the quantum numbers of a composite particle in terms of the quanum numbers of the preons of which it is composed. In the present case the vector of quantum numbers will have 9 elements corresponding to the 9 components of the 3×3 matrix. The three generations of a particle will have the same quantum numbers down the diagonal, and their off diagonal quantum numbers will be multiplied by cubed roots of unity.

Extending the Koide Relation to the Quarks

Unfortunately, the u, d, and s quark masses are not very well defined experimentally. The problem is that the masses and energies of objects built from these quarks are dominated by the off valence contributions. However, given that, it seems that a natural way of extending the Koide relation to the quarks is to apply MUB theory to the 3×3 matrices, that is, qutrits.

The MUBs of qutrits were discussed at length on this blog recently. One finds that one can define four bases. One of the bases will generate the 3×3 matrices used here. The other three seem similar to what one would expect with three colors of quarks, in three generations for a total of 3×4 = 12 particles.

However, the charged and neutral leptons do not share the same Koide formulas due to the presence of Berry phase in the neutrinos. To unify them with the quarks we first have to write a matrix that contains the charged and neutral leptons as well. So the actual number of particles we need is not 12, but instead is 24. This makes the whole idea rather muddy (somehow the MUBs for the 6-dimensional Hilbert space seem to come up) and so I doubt that MUBs have much to do with the generation structure of the quarks.

Another likely fruitful direction for research on Koide formulas are the weak mixing angles. For the leptons, these are particularly simple and appear to be, as one would expect, multiples of Berry phases between the MUBs of the Pauli algebra. In particular, note the use of the cubed roots of unity in the, frequently postulated neutrino mixing angles hep-ph/0202074.

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### 6 responses to “Koide formulas and Qubit / Qutrit MUBs”

1. Jay R. Yablon

Carl:

Enjoyed the fermion cube. Not too far from the multiplet you have seen at http://jayryablon.wordpress.com/2008/02/02/lab-note-4-an-interesting-left-chiral-muliplet-perhaps-indicative-of-preonic-structure-for-fermions/ . And, we are both scoping out preons.

Can you make that cube a product of simple gauge groups in some way? I went for SU(4)xSU(4)-bar, which is why I came up with a two-preon structure. Can you find some NxNxN structure to get the cube?

Also, Y and I^3 (my notiatiion) are on the plane of the page. What quantum number is on the “depth” on the cube going into the page? I found baryon number in my model.

You know that I have examined the possibility that the three quarks in a baryon emerge from non-Abelian (Yang Mills) magnetric sources. There is always the possibility that the same approach could be helpful to your quest.

Always a pleasure,

Jay

2. carlbrannen

Jay,

From the point of view of the preon theory I’m working on, the NxNxN structure should come from having each N be made up of two primitive idempotents taken from the complex Clifford algebra with signature (4,1).

If one tried to do this with the (3,1) Dirac algebra, one finds that one cannot get the right weak hypercharge and weak isospin numbers.

The restriction on the two primitive idempotents (that make up each N) is that they must share a certain eigenvalue that corresponds to velocity. The three N correspond to choices of the velocity in the x, y, and z directions.

In the usual Dirac complex Clifford algebra with signature (3,1), the operator for velocity in the z direction is zt or in gamma matrix form, $\gamma^3\gamma^0$. Similar for x and y.

So from a Clifford algebra point of view these are very simple things. If you demand that it be rewritten in symmetry form, it gets ugly. I guess the (4,1) Clifford algebra is represented by 8×8 matrices so the primitive idempotents are in an SU(8) representation. So the N are taken from SU(8) x SU(8), which is quite huge.

3. roger muldavin

Thought I ought to make a few quickie comments, at the least where I am on the subject:

{{Fig. 3: The fermion cube. The “Vr” is not shown for clarity}} and the (your link to “The Geometry 0f Fermions”) {{Fig. 1: Four deformations of space are shown}}

[Comments: Both Figures 1 and 3 are grids at right angles, but I have concluded that the grids if taken from three axis in space, Electromagnetic, these three would, with any wiggle room, orient in 3-D space at 120 degrees, that is, 2/3pi radians.

It seems to my visual mind that the grid for measurement, if it follows the “physics” would eliminate some variables.

Thus if Figures 1 and or 3 are projections of 3D space onto a flat surface, I have concluded that the Figure 1 grid ought to be the flat equal lateral triangle (felt), a regular triangle.

Thus, the flat equal edge rectangle (feer) if rotated so the top and the bottom vertices are at right angles to the felt base, gives a grid that, at the least, might help create a visual model, … and unify the grids with the physics.

The basic unit of the grid would then be either a rhombus or felt, the latter preferred since the felt can be the basic unit for constructing 3D polyhedra, a possible candidate for the Chemical Elements nuclii.

Hans Dehmelt proposed that the electron is a triplet of three vertices and three equal edges, he received a Nobel Prize circa 1998 for his work on isolating a single positron Penning Trap (one for over a month).

The Internet essays today shows, it seems, thousands of people working on this from the small laboratory experiments to the very large collaborative efforts requiring linear and circular accelerators.

I am struggling with as many models as my mind can tolerate, but the polyhedra as the pea in the middle of a football stadium and the electron in the bleachers, that is roughly the scale of the problem.

Consider that Calcium, Ca, if a polyhedron, has a 5-felt top, 10-felt ring middle, and 5-felt bottom.
Ca also has the smallest cross section to neutron collisions.

Additionally, to help me visualize such polyhedron constructions I purchased a larger set of the toy, Roger’s Connection (www.RogersConnection.com) that has as the basic magnetic units, vertices as steel balls some 1.5 cm diameter, and edges as a plastic tube about 10 cm length with small magnet set into each end to give a N/S direction to the whole edge, indented, with a diameter of some 0.6 cm. The indent appears to give a measurement between the center of the two steel balls when attached the plastic edge of very close to 12 cm.

The product literature states some 12 edges can be made, and that seems the case.

The use of this toy in conjunction with reading the “free” Internet essays has revealed to my mind the following:

(1) solid or stable polyhedra from as basic units that stand alone, that is appear solid when on a flat table, say Ns={1, 3, 4, 6, …}. This is tricky and needs more clarification; so next;
(2) unstable polyhedra placed on a flat surface,
say Nus={1*, 2*, 3*, 4*, 5*, …}; the “*” is to indicate that on a flat table the force of gravity can pull the steel balls downward, so this raises the issue of using the term “cap” and designating whether one counts the number of steel balls Nsb or the number of magnetized plastic tubes Nmt, further complicated or simplified at some more abstract level, that the vertice and the edge counts can be exchanged mathematically;
(3) So, rather than try the mathematical, pencil to paper other than making sketches in my notebook or at the edges of the B/W printouts, I just began constructions, the first successful, longest one, was a three rail ladder some ten links long for a full cycle, that is using felts I could construct a ten tetrahedronal helical 3-rail ladder, further more, it was hollow, and I mounted a penlight laser at one end and took a picture of the laser from the other end; my grandaughter about 3years of age when viewing the picture said: “what’s that light in the middle”;
(2) the idea of simply constructing single layered heights of tetrahedral, single tiers, then a larger five tier tetrahedron, I noticed there inside these large tetrahedron was “hollow” spaces; the I made a “Noah’s Arc” or “Fish” with open mouth and hollow spaces;
(3) After reading recent essays on porus materials, I noticed that four edged shapes, from orthogonal to non-orthogonal, gave spaces, as did penta-hedrons, and hexagons (6felts flat on table could be folded);
(4) The strength of larger shapes depended upon changing the N/S polarity of the edges on the vertices; Noah’s Arc and the Salmon Fish became very stable to lifting, the 10 section 3-rail, tetrahedron rail, hexical tube was not.

Enough for now, there is more, but in quick skimming of your great effort and results, there is a chance we are on the same pages.

Caps off,
hands on,
many models,
war goes,
peace on.

Best, rmuldavin