Edited 1/5/2018 4pm US EDST: At the bottom of this post, I have added an email to Dr. Klizting - I sent this email to him after he posted his answer to the question.


Imagine three sets of points P80, P32, and P64 so arranged that:

1) P80 and P64 together comprise the vertices of an object O144 with 144 vertices

2) P32 and P64 together comprise the vertices of an object O96 with 96 vertices

Is it possible to pick P80, P32, and P64 so that:

3) O144 actually contains two copies of the 72 roots of $E_6$ (two $1$$_2$$_2$'s)

4) O96 actually contains two copies of the 48 roots of $F_4$ (two 24-cells)

5) O144 and 096 together comprise one copy of the roots of $E_8$ (one $4$$_2$$_1$)

(Recall here that $E_6$ and $F_4$ are related by folding of Coxeter-Dynkin diagrams - see


https://en.wikipedia.org/wiki/F4_(mathematics) )

Thanks for whatever time you can afford to spend considering the possibility of this construction.

Email to Dr. Richard Klitzing 4 Jan 2018:

I want to thank you again for the description of the birectified hexeract which you gave at MSE.

Here is why it may prove to be both useful and important.

As I may have mentioned, my team is working at two biomolecular levels simultaneously:

1) the level of DNA and RNA polynucleotides and their associated energetics

2) the level of protein polypeptides (amino acid chains) and two of their associated properties (amino acid hydroaffinity and associated tRNA synthetase class)

In addition, because these two levels are interrelated by what is commonly called the "genetic code", my team is working at the junction of these two levels, i.e. the interface at which DNA genes are transcribed into RNA messages which are then translated into the polypeptide chains of protein "primary structures."

We are fairly certain at this point that at the level of polynucleotides, we can profitably make use of E8 to characterize a special set of 240 nonanucleotides (9-tuples over the DNA alphabet {t,c,a,g} or RNA alphabet {u,c,a,g}.

But these 240 nonanucleotide tokens at the DNA/RNA polynucleotide level translate (via the "genetic code") into 240 tripeptides at the protein polypeptide level, and it is quite possible that at this second level, the 240 tripeptide tokens can best be characterized in the way your birectified hexeract suggests.

And this possibility is all the more likely because your description of the birectified hexeract makes reference to penteract components, thereby invoking the 5-cube and 6-cube in a way which may well be related to the well-known fact that the genetic code itself "vacillates" between a 2^5 and 2^6 structure - to see this vacillation see this link (and the discussion of "Halitsky point-symmetry" therein:


Best regards as always ...

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    $\begingroup$ A short answer: No! The roots of $E_6$ and $E_8$ all have the same length. But the roots of $F_4$ are split into so called short and long roots - same number of each. This makes it trivially impossible to compose $E_8$ as a union in the prescribed way. $\endgroup$ – Jyrki Lahtonen Jan 4 '18 at 23:10
  • $\begingroup$ @JyrkiLahtonen - thanks as always! See Richard Klitzing's answer below, if you have not already seen it. Your answer reinforces the comment I made to him - that the empirically relevant symmetry structure may well involve $E_8$ at the level of polynucleotide energetics, but different symmetry structures at the level of the polypeptides encoded by polynucleotides via the genetic code. (continued next comment) $\endgroup$ – David Halitsky Jan 4 '18 at 23:17
  • $\begingroup$ @JyrkiLahtonen - in other words, the $E_8$ roots which I gave in this question math.stackexchange.com/questions/2578813/… are relevant to the polynucleotide level, but NOT to the polypeptide level or the encoding relation between the two levels . . . $\endgroup$ – David Halitsky Jan 4 '18 at 23:18

What me worries in your question, is the subsumed interplay of 4D (F4), 6D (E6), and 8D (E8) structures. Sure, there might be according subdimensional sections be involved, but these thereby would break down the total symmetry.

On the other hand it seems to me more that you come to those structures purely from the according numbers, applying those to those of vertices of respective polytopes or even more simply to the numbers of roots (the hull of which then would lead to these).

So I'd like to suggest to you a different toy:

Consider the 6D polytope o4o3x3o3o3o aka birectified hexeract. That one too would have a vertex count of 240. Moreover it allows a description as an axial stack (or lace tower description) as o4o3x3o3o (birectified penteract) atop o4x3o3o3o (rectified penteract) atop o4o3x3o3o again. All 3 vertex layers then will have 80 vertices each. Even more interesting to you would be the fact, that each of those in turn could be described as an axial stack. The o4o3x3o3o (birectified penteract) would give rise to the lace tower o4o3x3o (rectified 16-cell = 24-cell) atop o4x3o3o (rectified tesseract) atop o4o3x3o again. And o4x3o3o3o (rectified penteract) would give rise to o4x3o3o (rectified tesseract) atop q4o3o3o (sqrt(2)-scaled tesseract) atop o4x3o3o again.

Thus to conclude, one could divide the original o4o3o3o3o3o symmetry group into o4o times o4o3o3o, and thus display the original o4o3x3o3o3o (birectified hexeract) in an according projection as the lace city

i    r    i

r    q    r

i    r    i

where i = o4o3x3o = 24-cell with 24 vertices each, r = o4x3o3o = rectified tesseract with 32 vertices each, and q = q4o3o3o = sqrt(2)-scaled tesseract with 16 vertices.

What I'm describing here is just an according orientation of the vertex set of the original 6D polytope. It can be aligned such as to project onto 2D in the given lace city display. And the corresponding perp space projections then would be the vertex sets of the mentioned polychora.

That orientation now makes it easy to find all the numbers you were dealing with. E.g. just diminish that 6D polytope in that sense as to chop off the 24-cell projections at the corners. This then would reduce the original vertex count of 240 by 4 x 24 = 96. That is this diminishing then remains with 144 vertices. If you'd further omit the top and bottom r as well, that is you'd Thus delete the top and bottom vertex layer fully, then you are left with the medial one, which has a vertex count of 80.

For sure, there might be different constellations of points in space, which show up your numbers as well. I just aimed to provide an example, which does include all these numbers, but which does not take refuge to that cross-dimensional lunacy.

--- rk

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    $\begingroup$ Thank you for taking the time to consider the question, and also, of course, for the time you took to explain it clearly!!! Your suggestion raises a possibility that my teams's 240 tricodons (ordered 3-tuples of codons) can best be described ENERGETICALLY in terms of the structure of $E_8$, but by other structures for other purposes - e.g. their relation to the tripeptides (ordered 3-tuples of the amino acids) which they encode, and the properties of those 3-tuples related to amino acid hydroaffinity and synthetase class. (continued next comment) $\endgroup$ – David Halitsky Jan 4 '18 at 23:03
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    $\begingroup$ Regarding our energetic characterization of tricodons in terms of $E_8$, if you have any thoughts on the nearest-neighbor relation requested here math.stackexchange.com/questions/2578813/…, I would of course be grateful to know them. (continued next comment) $\endgroup$ – David Halitsky Jan 4 '18 at 23:05
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    $\begingroup$ Finally, I am curious to see if Wendy comes up with yet another alternative, since you've mentioned that others might well exist. $\endgroup$ – David Halitsky Jan 4 '18 at 23:06
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    $\begingroup$ @David I really think that this game would be easier if you described the tricodons as vectors somehow rather than as strictly combinatorial objects. At the beginning I expected you to point us to some source describing some physical/chemical/geometrical properties of the relevant molecules, and then could start studying what kind symmetries are present in that data. If it turns out there are enough "reflection" symmetries (where a "reflection" may also be a rule to replace one codon with another), then we could figure out which group that is. It might even be $W(E_8)$. $\endgroup$ – Jyrki Lahtonen Jan 4 '18 at 23:24
  • $\begingroup$ But I'm afraid this looks more and more like random poking. $\endgroup$ – Jyrki Lahtonen Jan 4 '18 at 23:25

There is a projection of an eight-dimensional lattice onto 4BB, and with a little packing, we can get a direct packing of E8 onto 4D. 4BB is the lattice whose eutactic star is two reciprocal {3,4,3}, of the same size. It is infinitely dense, but piecewise discrete. There is a similar kind of packing of E8 into {5,3,3,5/2} too.

It would require that there is a second 4BB, say 4BB', which corresponds to a 1-1 mapping of E8 into four dimensions. I have not found any regular source of mathematics on it, so i rely mostly on my own diggings. 4BB and 4BB' are Z4 systems, that is, the integer span over Z[1,$\sqrt{2}$]. {5,3,3,5/2} is one of a dozen known primitives for the Z5 = Z[1,ø] system.

  • $\begingroup$ @wendykrieger - Thanks for this addditional perspective on the question ! $\endgroup$ – David Halitsky Jan 6 '18 at 12:36

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