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If you look at the section "Roots of E6" here:

https://en.wikipedia.org/wiki/E6_(mathematics)

you will see that ALL the given 9-tuples of coordinates have internal triplet structure (to see this in the first 18 9-tuples, simply pay close attention to the difference between semicolons and commas in these 18 9-tuples.)

So, my question is why the most symmetric coordinatization of the 72 roots should have this uniform triplet structure?

Does it "just so happen" to work out this way, or is there actually some reason why this triplet structure must occur ?

Please note that I'm looking for purely mathematical answer here, such as "you can't get perfect symmetry across 72 points in 9-space without somehow inducing triplet structure", NOT an answer that makes reference to any instantiations of the group, e.g. in particle physics, etc.

And thanks for whatever time any one can afford to spend considering this question.

11/18/2017: Here is a partial answer to my question, again from Dr. Richard Klitzing.

We have the "exceptional" 18 vertices as:

(1,−1,0;0,0,0;0,0,0), (−1,1,0;0,0,0;0,0,0),
(−1,0,1;0,0,0;0,0,0), (1,0,−1;0,0,0;0,0,0),
(0,1,−1;0,0,0;0,0,0), (0,−1,1;0,0,0;0,0,0),
(0,0,0;1,−1,0;0,0,0), (0,0,0;−1,1,0;0,0,0),
(0,0,0;−1,0,1;0,0,0), (0,0,0;1,0,−1;0,0,0),
(0,0,0;0,1,−1;0,0,0), (0,0,0;0,−1,1;0,0,0),
(0,0,0;0,0,0;1,−1,0), (0,0,0;0,0,0;−1,1,0),
(0,0,0;0,0,0;−1,0,1), (0,0,0;0,0,0;1,0,−1),
(0,0,0;0,0,0;0,1,−1), (0,0,0;0,0,0;0,−1,1),

Each block of 3 rows, i.e. 6 of those 9-tuples, can be seen to host its non-zero values within a single triple, that is those live in 3 mutually perpendicular subspaces. But 6 (the true embeding space of mo) : 3 (perp spaces) = 2 (dimension of each).

Considering furthermore that the radius of mo equals 1 edge length unit, we thus have to scale those 9-tuples by a factor of 1/sqrt(2). But then the distance between any tuple and one of a different row (of the same block) likewise would yield either a value of 1 or of sqrt(3). The distance between the 2 tuples of a row then becomes 2.

Thus indeed, those 3 blocks of 3 rows each define a planar regular hexagon each.

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  • $\begingroup$ Dr. Richard has provided at least a partial answer to this question, and I've added it to the main post itself. $\endgroup$ – David Halitsky Nov 18 '17 at 21:17
  • $\begingroup$ Working on an explanation in terms of the related Coxeter group and its subgroups. This will take a while. It is easy to describe those 18 roots with single non-zero triple using the Dynkin diagram and "standard" facts about root systems. Understanding the roots with non-zero coordinates is (as you hinted in another question) probably related to projections that I should also be able to describe. I know you have moved forward already - I'm looking at this to satisfy my own curiosity :-) $\endgroup$ – Jyrki Lahtonen Jan 15 '18 at 10:14
  • $\begingroup$ @JyrkiLahtonen - although we have moved forward, this question is still very important because of some new results which Will Orrick and I are now seeing in regard to cyclic permutations of 9-tuples such as 456789123 and 789123456 when applied to 9-tuples over 2-letter alphabets. Also, Richard Klitzing and Wendy Krieger have together provided two pieces of information which I think you may find of general interest. (see next two comments for Richard's info and Wendy's info) - I cannot begin to tell you how excited I am about both their observations. $\endgroup$ – David Halitsky Jan 15 '18 at 14:55
  • $\begingroup$ @JyrkiLahtonen here is Richard's new piece of info: Regarding the dissection of the vertex set of the Gosset polytope 4_21 according to 240 = 84+72+84, this decomposition is based on the fact that 4_21 can be described as the convex hull of 2 mutually inverted copies of a birectified enneazetton (each 84 vertices) and one expanded enneazetton (72 vertices). (Note here that an "enzeatton" itself is simply an 8-simplex.) $\endgroup$ – David Halitsky Jan 15 '18 at 14:56
  • $\begingroup$ @JyrkiLahtonen - here is Wendy's new piece of info: The largest shared set between A8 (ie the 72 in the {84,72,84}), and E6, is A5 (30 vertices). The remaining 42 vertices (roots of E6) are spread equally between the two sets of 84 in the {84,72,84} decomposition. $\endgroup$ – David Halitsky Jan 15 '18 at 14:58

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