The generalized Kronecker delta and three sets of 16 tetrahedra defined by 192 of the 240 roots of E8 (vertices of Gosset's 8-polytope 4_21)

Original question (without additional information from Wendy):

Using 192 of the 240 roots of E8 (vertices of 4_21), Wendy Krieger has defined 48 disjoint tetrahedra this way:

Taking the E8 as {128,112}, of radius 2, we get

16 tetrahedra at (1,1,1,1)E, (1,1,1,1)E

16 tetrahedra at (1,1,1,1)O (1,1,1,1)O

16 tetrahedra in (2,0,0,0)A (2,0,0,0)A

In the first two,

E means take an even number of sign-changes in the bracket.

O means take an odd number of sign-changes in each bracket.

A means all permutations, all change of sign in the brackets.

The vertices of the tetrahedron then comes from three coordinates in a given set, so these are the coordinates of a tetrahedron in the first set, using the first three coordinates.

1,1,1,1 1,1,1,1

1,-1,-1,1 1,1,1,1

-1,1,-1,1 1,1,1,1

-1,-1,-1,1 1,1,1,1

Question:

Is it possible to define a generalized Kronecker delta function which takes Wendy's E, O, and A sets to 1, -1, and 0 respectively (or -1, 1, 0)?

See this link for definition of the GENERALIZED Kronecker delta:

https://en.wikipedia.org/wiki/Kronecker_delta

New information from Wendy Krieger (added 9/18):

Of the 240 total roots of E8, 192 are consumed by Wendy's 48 tetrahedra, leaving 48. And she has discovered that these 48 define two 24-cells. So in addition to the generalized Kronecker delta question, there is an additional question as to whether the 48 vertices of the two 24-cells "organize" the 48 tetrahedra in any interesting way.

Wendy has provided an affirmative and interesting answer to this question:

The centres of the 48 octahedral faces of each of the two 24-cells, in rectangular product, produce a 3*3 array of the 48 tetrahedra, of which there are six distinct sets of 24.

See this link for definition of the famous and unique 24-cell, which has no analogs in spaces lower or higher than 4 dimensions:

New information from Dr. David Richter (9/24/2018)

Dr. Richter has kindly suggested that Wendy's construction is not new within the literature of Lie-algebras. He writes:

"It does not seem new to me. Eugene Dynkin studied these things in depth in the 1940's, for example. (In Russia during World War II. He avoided the draft due to poor eyesight.) You should look up his article "Semisimple subalgebras of semisimple Lie algebras". This was published in the Translations of the American Mathematical Society in 1952. Although he does not use the language of regular polytopes, I think you will find the structures that you describe, encoded as rank -8 Lie subalgebras of the E(8) Lie algebra."

New information from Dr. Derek Smith provided 10/1/2018)

There are 16 "nearest neighbors” to any one of Wendy's tetrahedra, in the following sense.

Consider the tetrahedron T that has its four vertices in R^8 given by the following rows (here + is 1, - is -1, and we’re using the even coordinate system, scaled up by a factor of 2 to avoid some fractions, so minimal non-zero vectors have norm 8):

+ + + + + + + +
+ - - + + + + +
- + - + + + + +
- - + + + + + +

The center c of T is the average of those vectors, namely c = 0 0 0 + + + + +.

The vectors in E8 closest to c are the four vertices of T, each of whose (squared) distance from c is 3. The next possible distance is 5, and the 16 vectors in E8 that achieve this are of the form 0 0 0 a b c d e, where each of the five letters is either 0 or 2, and there are an even number of 2’s.

As we’ve discussed previously, there are geometric symmetries of the lattice taking any one of the tetrahedra to any other, so there’s nothing special about this one. For instance, if we had taken

2 0 0 0 2 0 0 0
0 2 0 0 2 0 0 0
0 0 2 0 2 0 0 0
0 0 0 2 2 0 0 0

as our tetrahedron instead, with center c = 1/2 1/2 1/2 1/2 2 0 0 0, then the 16 vectors at distance 5 from c aren’t as easily described in one sentence; but you can check that they are

0 0 0 0 0 0 0 0
0 0 0 0 4 0 0 0
+ + + + + a b c   a, b, and c are + and -, with an even number of -‘s
+ + + + 3 x y z   x, y, and z are + and -, with an odd number of -‘s
0 0 0 0 2 p q r   p, q, and r are -2, 0, and 2, with exactly two 0’s

for a total of 1+1+4+4+6=16 vectors.

There may be other ways to describe “nearest neighbors” that might yield different answers, e.g. take vectors close to the 4 centers of the tetrahedron’s faces, or close to the 6 centers of the tetrahedron’s edges.

Email from Roger Bagula 10/21/2018 (and mine to him):

My email to Roger Bagula (10/20/2018):

My research team seems to have uncovered a pretty clear biomolecular
instantiation of OEIS A152459.
Furthermore, this instantiation appears to be connected with a particular
decomposition of the root-system of E8 (via the geometry of 4_21 or
a Dynkin sub-algebra).

His email to me (10/21/2018):

Thank you for letting me know about your discovery.
That idea / result appears to be an amazing connection of SL(2,C) type
of 2x2 matrices to the higher geometry.
I wish you luck as E_8 symmetry breaking is near my heart, LOL.

Email from Jim Humphreys (10/21/2018) re "SL(2,C)"

Probably the quoted phrase just refers to the group of complex 2x2
matrices of determinant 1, usually referred to as the (Lie or linear
algebraic) group SL(2.C).
Jim Humphreys

Email from Roger Bagula 10/22/2018 with some further thoughts

This morning I was thinking about E_8 from the Thurston-Weeks minimal
hyperbolic 3 manifold approach.
The original Poincare dodecahedral universe model is elliptical,
but the Siefert-Weber dodecahedral space is hyperbolic.
Weeks contends that the Cosmic microwave background radiation should be
invertible as an X-ray crystal is, so the the spectrum of anomalous lines
would reflect the geometry of the universe that contains them. (That is
looking at the bare wiggles on the blackbody peak /anisotropy.) The Fourier
transform inversion approach suggests itself.
https://www.physicsmyths.org.uk/imgs/wmap_spec.gif

A_5 as a 3d point group is Ih (Icosahedral group) as 60 elements, so the 240
of your 8 dimensional model (4_21) is 4 orthogonal Icosahedral groups as a
matrix group:
{{A_5,A_5},
{A_5,-A_5}}
In trace-Character terms that looks very much like an Hadamard matrix self-
similar construction, as the  Ih :A_5 character table is already an
{{C,C},
{C,-C}}
symmetry.
In spectral terms this situation is a good thing as only a limited number of
vibrations are spectrally active in the Infrared . Weeks in one of his
papers points out the observed spectra is more limited than expected from
the hyperbolic 3 manifold symmetry models available.
The C60 buckyball is an elliptical molecular analog to this sort of Ih
symmetry: (with a spectral list of frequencies)
http://www.public.asu.edu/~cosmen/C60_vibrations/newc60revcorr.pdf
The Raman shift in the 1466/cm to 1474/cm region looks like a backward
cosmic microwave background anisotropy.
There are actually very few symmetry models that would be able to have such
a limited spectral set in the microwave region. E_8 as an Ih ( A_5)
hyperbolic 3 manifold is one of them.

Added 10/28/2018 - a possible different way to derive Wendy's sets of 16 tetrahedra within the 4_21 polytope:

Recall that inside the 4_21 polytope which instantiates the roots of E8,
Wendy's 3x3 matrix (derived from two 24-cells in the 4_21) generates
48 tetrahedra inside the E8 (in six different ways), and that
each such set of 48 divides naturally into three sets of 16 tetrahedra.

Nature is telling us very loudly and clearly  that it must be possible
to take the sets of 16 tetrahedra in sets of five in some reasonable way,
and further, that it must be possible to take four such sets
of eighty tetrahedra in some reasonable way.

So, can anyone think of any reasonable geometric, lattice-theoretic,
or group-theoretic way to get 5*16 = 80 and 4*80 = 320,
using the various sets of 16 tetrahedra generated by Wendy's construction?

Roger Bagula has just reported that the group SO(27) appears to be occurring within our biomolecular instantiation of the "Krieger-tetrahedra" in 4_21. This may be of possible relevance since 27*26 = 702, where 702 is the number of 4-faces of 1_22 (which realizes the 72 roots of E6 within the 240 roots of E8 realized by 4_21.)

An important piece of Mathematica code from Roger Bagula (important because the value 351 and the triple of absolute values |1,14,25| occur in an obviously related way within the energetics our biomolecular instantiation of Wendy's "Krieger-tetrahedra".)

A linear relationship between {1,14,25} and {-351,351}:
(* Mathematica*)
(*diff vector*)
v = {1, 14, 25}
{1, 14, 25}
(* plane vectors*)
w1 = {-1, x, y}
{-1, x, y}
w2 = {1, y, x}
{1, y, x}
(*linear solutions*)
NSolve[{v.w1 - 351 == 0, v.w2 + 351 == 0}, {x, y}]
{{x -> -32., y -> 32.}}
(*Check*)
-1*1 - 32*14 + 32*25
351
1 + 32*14 - 32*25
-351
(*end*)
32=16*2

The SO(27) group:
number of elements; n(n-1)/2=27*26/2=351

Added 11/4/2018: i) an email from Dr. Richard Klitzing describing 40 pairs of pentachora within 1_22 (the polytope which realizes the roots of E6); and ii) an email from me to Richard and Wendy Krieger conveying a possible reason why our biomolecular energetic data might be showing not only patterns related to Wendy's tetrahedra within 4_21, but also Richard's pentachora within 1_22.

Email 11/4/2018 from Dr. Richard Klitzing:

In my prior email, I described a setup of 40 pairs of pentachora
(i.e. tetrahedral pyramids), being pairwise connected at their tips,
which are to be found within 1_22. And that this number of 40 furthermore
is being divisable as 40=1+1+2+12+12+12.

Btw. this can be depicted in the following lace city display
(aka projection of vertex set onto 2D) of mo = 1_22,  which projects along o
rhtogonal A_3 x A_1 symmetry:

a   B   a            -- x3o3o *b3o3o (hin)

o   O   c   O   o        -- o3o3o *b3x3o (rat)

b   A   b            -- o3o3x *b3o3o (alt. hin)

|   |   |   |   +-- o3o3o3o3o (point)
|   |   |   +------ o3o3x3o3o (dot)
|   |   +---------- x3o3o3o3x (scad)
|   +-------------- o3o3x3o3o (dot)
+------------------ o3o3o3o3o (point)

where:
o = o3o3o o (point)
a = x3o3o o (tet)
b = o3o3x o (dual tet)
A = x3o3o x (tepe)
B = o3o3x x (inv. tepe)
O = o3x3o x (ope)
c = compound of
x3o3x o (co)
o3o3o u (ortho u-line)

Email 11/4/2018 from me to Dr. Richard Klitzing and Wendy Krieger:

This is just a note to convey my own personal suspicion as to why
our biomolecular energetic data seem to be showing internal structure
related to pentachora within 1_22 as well as tetrahedra within 4_21
(where 1_22 and 4_21 respectively realize the roots of E6 and E8.)

On the empirical biomolecular energetic side, we calculate energetic
values for certain linear ordered 6-tuples, where these 6-tuples
come in natural sets of 64 that can be placed into correspondence
with the 64 vertices of one of Wendy's sets of 16 tetrahedra within 4_21.

But to calculate an energetic value for any ordered 6-tuple (1,2,3,4,5,6)
in our various sets of 64 6-tuples, our protocol demands that we decompose
the 6-tuple into five overlapping 2-tuples (12,23,34,45,56), in accordance
with the analysis in the seminal paper by Jacques Fresco
(the senior scientific member of our team, now Emeritus Princeton):

https://www.ncbi.nlm.nih.gov/pubmed/6193821

So, if we take eight of our biomolecular 6-tuples that correspond to two
of Wendy's tetrahedra in 4_21, these eight will contain 40 of our
(overlapping) 2-tuples, and this is why I think that our data are
ALSO showing patterns pointing to the pentachora within 1_22, i.e.
patterns involving the triples of integers (1,14,25) in relation to
the integer 351.

For this reason, it would be very nice if some regular relationship
could be shown between the tetrahedra in 4_21 and the pentchora within 1_22.

But as Wendy has pointed out, it is not a trivial task to determine
whether any such relationship exists, and if so, what this relationship is.

In this regard, Roger Bagula and our team's statistical guru Ray Koopman
are now doing further numerical analyses of our (1,14,25) patterns in
relation to our (351, 702, 1404) patterns, and perhaps these analyses
will provide some clues as to how the pentachora in 1_22 might lie
"nicely" with respect to the tetrahedra in 4_21.

But again, all of the above is nothing more than a suspicion, at least
at this point.

Added 25 November 2018

For those who find this thread of interest, here is something important to remember. Wendy defines a 3x3 matrix of 16 cells, each of which is the dual of a 4-cube (tesseract). So when we select any set of three 16-cells from this matrix choosing one cell from each row and one cell from each column of the 3x3 (e.g. cells 11, 22, 33), we are implicitly choosing three 4-cubes - namely the duals of the three 16-cells which we have selected. And if we then use Coxeter's standard orthogonal projection of 4-cube to rhombic dodecahedron (in which two vertices of the 4-cube go to the center of the rhombic dodecahedron), some very nice things happen on the scientific side of the house, particularly if we make the slightly unusual choice to use Coxeter's projection so that the vertices 1010 and 0101 becomes the poles of the dodecahedron (instead of 1111 and 0000, which are the usual choices.)

Here is a correction by Wendy Krieger to my comment just above.

From two given 24-cells which are orthogonal, you can make nine bi-16ch
prisms.  Any three of these that are not in the same row or column make
a 4_21.

Each of the bi-16chora, make 16 tetrahedra, which (supposing some
four-axis * four-axis), lie by pairs at the vertices of a perpendicular
16-choron.  What this means, is for example, an octahedron can be
regarded as a pair of triangles opposite each other.  If this octahedron
lies in a prism-product with something else, then each vertex of the
'something else' gives an octahedron, and thus a pair of opposite but
parallel triangles."

So, each set of "Krieger-tetrahedra" actually defines eight 16-cells, each of which defines eight dual 4-cubes which can each be projected into a rhombic dodecahedron in three of its four dimensions.

See the post by Dr. Richard Klitzing with time stamp "Sun Dec 09, 2018 2:58 pm" in this thread at the "hi.gher.Space" bulletin board:

http://hi.gher.space/forum/viewtopic.php?f=25&t=1881

Richard has succeeded in understanding and documenting the full structure of the scaliform polytope "codify", which is his name for the polytope with 192 vertices derived from 4_21 by Wendy's construction (in which she removes the 48 vertices of two 24-cells from 4_21.)

• It would he helpful for Lie theorists to give some definitions of symbols used here (and to add more tags). Note too that the $E_8$ root system (W. Killing, E. Cartan) has 240 roots in all, half being positive. – Jim Humphreys Sep 16 '18 at 22:43
• @JimHumphreys - yes - 240 in all. And with respect to this 240 - permit me to mention something which I didn't state in the question. Of the 240, 192 are consumed by Wendy's 48 tetrahedra, leaving 48. And she has discovered that these 48 define two 24-cells. So in addition to the generalized Kronecker delta question, there is an additional question as to whether the 48 vertices of the two 24-cells "organize" the 48 tetrahedra in any interesting way. Finally, I have edited the title to make it clear that I'm speaking of 192 of the total 240. Thanks for taking the time to comment. – David Halitsky Sep 16 '18 at 23:02
• David, all those "Edited" notes at the beginning are unnecessary (all the revisions are already archived in case anyone wants to look them up), and they get in the way of getting to the actual question. In general, while we want precision, and we generally welcome some setting of context, we should avoid overbloated questions -- and this one runs that risk, even after I edit out the "Edited"s. – Todd Trimble Oct 28 '18 at 15:18
• Most likely I am wrong since I don't actually understand almost everything you write, but - product of all coordinates is 1 for E, -1 for O and 0 for A. Is this what you want or? – მამუკა ჯიბლაძე Nov 4 '18 at 19:24
• Given that this question has an answer that you find acceptable (and you can "accept" your answer by clicking the check mark), further edits to the question seem to be unnecessary. I suggest you add any additional news in the form of comments on the accepted answer. – S. Carnahan Dec 10 '18 at 11:59

Based on the comments from Wendy Krieger below, I accept this observation from მამუკა ჯიბლაძე as an acceptable answer to the question

"product of all coordinates is 1 for E, -1 for O and 0 for A"

Comments from Wendy Krieger 11/06/2018:

There is a geometric description here, which has to deal with
reflections.  A reflection in a mirror will produce a reversed image,
that is, convert 1 to -1.

If you place two mirrors at right-angles, the image in the corner is a
double reflection, or -1 * -1 = +1.  So if you looked at yourself in
this arrangement, when you lift your right hand, the image lift its
right hand too.  In a single reflection, a right hand is reflected to
the image's left hand.

The most likely source for this is the rectangular mirrors (the ones
like x=0, and y=0, and z=0), are flipping the image back and forward.
What it probably means in terms of your experiment is that A-X and A+X
somehow add to the same measure, but the difference from A is being
inverted.

The whole of E8 is a eutactic star.  The two 24-cells are eutactic
stars, severally and jointly, and the thing you are looking at is one of
the 16chora inscribed in the 24-cells.

The 24ch is the group of order 192, can be divided into three 4r's
(16ch) or lines at right angles.  These three can be labeled say N, E,
O.  The vertices of the form (1,1,1,1) etc, can be formed by an 'even'
sum of four N axies, or an odd sum of four O axies. even and odd here
means that the axies are labeled as +1 to -1, and an e/o number of
negatives are used.

Likewise, (2,0,0,0) can be formed by an 'even' number of E or an 'odd'
number of O vectors.

The signs then correspond to reflections in the N star, which inverts
exactly one axis (eg w,x,y,z -> -w,x,y,z) and to get to other points,
you have to reflect in the x, y, and/or z mirror too.

The progression from O to N to E is a linear operation in odd
dimensions, but involves some sort of turn in even ones.  This is why
they can be treated symmetrically in 4 and 8 dimensions.

Added 20 Dec 2018:

For those familiar with at least the rudiments of biomolecular translation (the process by which messages transcribed from genes are translated into proteins), the following conceptual cheat-sheet may help explain why an affirmative answer to the "generalized Kronecker delta" question was so important to my research team. In order to be able to deliver on the claims made in this cheat-sheet, we needed to be able to show that given Wendy's construction, we can define a "high" (+1), "low" (-1), and a midpoint between them (0). And the affirmative answer provided to the question by @მამუკაჯიბლაძე tells us that this is possible.

Conceptual Cheat-Sheet

Our analysis deals solely with sets of dicodons and the sets of dipeptides
they encode, NOT with individual codons and encoded amino acids, nor with
individual dicodons and encoded dipeptides.

Our analysis deals solely with the energetic properties of sets of dicodons,
the affinity properties of sets of dipeptides (hydrophobicity), and the
synthetase affiliation properties of sets of dipeptides (Class I or Class
II).  No other properties of dicodons or dipeptides are relevant to the
analysis.

Our analysis identifies certain energetic symmetries in the energetic
patterns exhibited by our sets  of dicodons.

Our analysis identifies certain affinity symmetries in the affinity patterns
exhibited by our sets of dipeptides.

Our analysis identifies certain affiliation symmetries in the affiliation
patterns exhibited by our sets of dipeptides.

Our analysis identifies certain consistent symmetry relations between
energetic symmetries and affinity symmetries, and also between energetic
symmetries and affiliation symmetries.

These symmetry relations hold for: i) both amino acids of dipeptides (AA1
and AA2 in a dipeptide AA1AA2); or ii) AA1 only; or iii) AA2 only; or iv)
neither.  And we interpret (i-iv) as suggesting that sets of dipeptides
assumed functionality in protein structure according to this rough 3-way
chronology:

Early onset of functionality:   sets of dipeptides exhibiting symmetry
relations to their dicodon sets for both
AA1And AA2
Late onset of functionality:    sets of dipeptides exhibiting symmetry
relations to their dicodon sets for
neither AA1 nor AA2
Onset midway:                   sets of dipeptides exhibiting symmetry
relations to their dicodon set for AA1
or AA2 but not botj

Our rationale for this chronology is that early onset dipeptide sets assumed
functionality when mRNA energetics were still important in early translation
systems, whereas late onset dipeptides assumed functionality in relatively
mature translation systems in which mRNA energetics were relatively less
important (primarily due to the advent of the water-tight ribosome.)

This hypothetical chronology is fully falsifiable by determining if it makes
correct predictions with respect to “early-late” pairs of SCOP protein
families within SCOP superfamilies (where protein family age is taken as the
ranking assigned by GCA and his team.)

Since the 72 vertices of the polytope 1_22 occur as 72 of the 240 vertices of the polytope 4_21 (corresponding to the fact that the roots of E8 contain a copy of the roots of E6), Dr. Richard Klitzing kindly investigated whether there might be an empirically relevant construction inside 1_22 which might be intrinsically related to Wendy Krieger's empirically relevant tetrahedral construction within 4_21. Richard has determined that in 1_22, one can find 40 pairs of opposed pentachora (aka 5-cells, aka hypertetrahedrons in 4-space) and these 40 pairs seem to be empirically relevant to our biomolecular results. However, Richard and Wendy have not yet determined whether Richard's pentachora are related to Wendy's tetrahedra in any meaningful way. Also, note that this question is a new particular version of the more general question asked in this post last year: $E_6$, $E_8$, and Coxeter's (anti-)prismatic projections of the n-dimensional cross-polytopes

Here are further details on the pentachora from Richard, in reponse to my question to him as to whether his 40 pairs of pentachora can be divided into three subsets of 10, 20, and 10.

Those 2 x 40 pentachora within that E_6 figure 1_22 (= mo) had been obtained
from its D_5 subsymmetry, when being seen as hin (= 1_21, hemipenteract)
atop rat (rectified 5D crosspolytope) atop alternate hin (i.e. the “other”
hemipenteract) as xoo3ooo3oox *b3oxo3ooo&#xt. In fact, each hemipenteract
layer shows up 40 tetrahedra which are solely connected to hexadecachora
(within that very layer). (In fact, those are the following tetrahedra:
x3o3o *b3.3.) Those tetrahedra furthermore connect to the medial layer
(featuring exactly 40 vertices) via tips, thus becoming pyramids with
tetrahedral bases, aka the mentioned pentachora.

Now you ask whether those 40 pentachora (on either side) would be divisable
into a sum of 10+20+10. Thus your question readily is transfering down one
dimension, yielding the quest to divide those 40 tetrahedra within either hin.

And indeed, when considering hin in turn as an axial stack (aka lace tower)
of a point atop a rectified penteract atop a (relatively inverted) penteract,
i.e. in axial A_4 symmetry, as ooo3oxo3ooo3oox&#xt, then those 40 tetrahedra
show up as

10x  oo.3ox.3...3...&#xt
20x  ...3.xo3.oo3...&#xt
10x  ...3...3.oo3.ox&#xt

which is nothing but a positive answer to your quest.

Thus in total you break the E_6 symmetry down into a lace city (i.e. a 2D
position space) with a 4D perpendicular object space. That perp space then
still features A_4 symmetry. (Btw., the corresponding position space
geometry is readily visible displayed as the 2nd provided lace city of
https://bendwavy.org/klitzing/incmats/mo.htm.)

And here is my response to Richard:

Thank you very much, Richard!

I am very glad that there is an affirmative answer to the question,
because it arose from a consideration of this empirically derived table:

Min
Asym    2       1       0

yr>ry   1       5       3        9
yr=ry   8       9       2       19
yr<ry   1       6       5       12
10      20      10       40

I won't bother at the moment to tell you what the counts in this
matrix are - it's too complicated for a short email.)

But I do want to draw your attention to the fact that the 3x3 matrix here
is our empirical counterpart to Wendy's 3x3 matrix from which her three
sets of 16 tetrahedra can be selected in 6 different ways.

And in my own personal opinion, the fact that we get these 10-20-10
column sums strongly suggests that your pentachora in 1_22 (E6)
can in fact be systematically related to Wendy's tetrahedra in 4_21 (E8),
assuming of couse that we choose to locate the1_22 inside a 4_21
(i.e. to locate a copy of the roots of E6 inside the roots of E8.)

Note that in the 3x3 immediately above, these two 2x2's occur

5 3
9 2

9 2
6 5

where: i) 5*2 = 10, 3*9 = 27, and 10*27 = 270 ii) 9*5 = 45, 2*6 = 12, and 45*12 = 540 iii) 2*270 = 540.

Coincidence? Perhaps. Perhaps not.

Added 1/5/2019 as a more sensible refinement of the above note:

Consider any three 3x3's:

SM:
sm11 sm12 sm13
sm21 sm22 sm23
sm31 sm32 sm33

SD:
sd11 sd12 sd13
sd21 sd22 sd23
sd31 sd32 sd33

MD:
md11 md12 md13
md21 md22 md23
md31 md32 md33

which exhibit the triple cross-ratio:

(sm12*sm13)/(sm32*sm33) = (sd12*sd13)/(sd32*sd33) = (md12*md13)/(md32*md33)

Our empirical data exhibit three such 3x3's:

Min
2      1      0
yr>ry  1      5      3    9
yr=ry  8      9      2   19
yr<ry  1      6      5   12
10     20     10   40

Cross-Diff
1     14     25
yr>ry  3      3      3    9
yr=ry  7      6      6   19
yr<ry  3      6      3   12
13     15     12   40

Cross-Diff
1     14     25
2      5      1      4   10
1      7      6      7   20
0      1      8      1   10
13     15     12   40

and it will therefore be interesting to see if Richard Klitzing can see a geometric realization of this linear-algebraic cross-ratio in his construction involving 40 pairs of pentachora within 1_22 (roots of E6.)

Added 25 January 2019:

I asked Wendy Krieger this question:

> Recall that we have 3 distinct sets with 16 Krieger-tetrahedra each,
> for a total of 48 distinct tetrahedra that consume 192 vertices of the
> 4_21.
>
> How many distinct 16-cells can be found among these 48 tetrahedra (in
> various ways, of course)?
>
> Please clarify when you have a moment - thanks.

The 48 distinct tetrahedra are opposite faces, by pairs, of 24 16-cells:

192 vertices -> 3 * (16ch * 16 ch)-prisms = 3*8*8 = 192.

At each prism, we take each vertex of the first 16ch, * a 16ch, gives 8
16ch per prism.

Each 16ch gives two opposite tetrahedra.

So there are 24 such 16ch.

Added 27 Jan 2018

I asked Wendy Krieger this question:

Wendy has told us that any set of 16 Krieger-tetrahedra is spanned by
eight 16-cells.  (Each of these 16-cells pairs-off 2 of the tetrahedra.)

By duality, these eight 16-cells define eight tesseracts with a total
of 128 vertices.

But how many distinct vertices among these 128?

Is it easy for you to answer this question?

If so, then depending on your answer, here is my next question.

How many distinct penteracts can we find in the distinct vertices of
the 128?

The questions are both easy.

If you take your 4_21 as having coordinates at OO, EE, NN, then the dual
tesseracts will form at places like OE+ON, EO+EN, NO+NE.  None of these
are in the 4_21, but alternate vertices lie in the other 4_21's that
form in the 3*3 grid.

There are 48 such tesseracts, that form into a chain of six, that is,
you get a skew hexagon formed by a ring of six tesseracts joined at the
opposite cubes, eg OE-ON-EN-EO-NO-NE-OE...

The vertices of these 48*16 = 256, lie in two sets of 192 vertices of
each of the two E8s.  Because of this, one can replace the eight cubes
with a chain of eight 16-cells that follow the same pattern.

The alternation of tesseracts can be seen from arranging a hexagon of
matches, and removing every second one.  But this still preserves the
384 vertices of two sets of 192, but each of the 24 remaining tesseracts
are un-connected.

There are no distinct tesseracts in this assembly, that i have seen.

Her answer here tells us how any choice of 48 Krieger-tetrahedra in a 4_21 are co-located with various sets of 80 Klitzing-tetrahedra in 1_22.

And this answer gives us an answer to the question I originally posed in this thread:

$E_6$, $E_8$, and Coxeter's (anti-)prismatic projections of the n-dimensional cross-polytopes

See new answer to this question added today.

Added 28 January 2019 - further notes from Wendy Krieger

The model i have been working with is D4D4. D4 equates to a form of
body-centred tesseract tiling.

D4 has four 'stations', which equate to where the D4 is sitting, along
with three seperate D4 cells in the centres of the three different
orientations of the 16ch that make this lattice.

When we take this D4D4 as an eight-dimensional lattice, the three
remaining sets of deep holes in square product form the three-by-three
matrix, from which the various E8's might be drawn.

E6 has a construction in D4H2 (H = hexagonal lattice)

If one takes a hexagonal lattice, it is possible to colour cells that
are on opposite sides of a neighbour in the same colour.  This results
in a four-colouring of the hexagonal lattice, which we label X, O, E,
N.  In D4H2, there is a distribution of 2_21, where the X-cells contain
just one vertex, and the O,E,N contain 8 vertices.  Given a vertex in
any of the four colours of H2, we get the other three cells representing
a distribution of 16ch in the same arrangement.

Since the H2 maps onto D4 as a van Oss polygon, it comes that diameters
of the 3,4,3 give rise to a hexagon as described above, and then there
are 16 such girthing hexagons on the {3,4,3}, that are notionally in a
group AA4 (bi-alternating group 4), represented by the even alternation
of rows+columns of a 4*4 matrix.  Any row or column represents a set of
four hexagons that cover the full hexagon

At any point in E8, it is thus possible to have four crossing E6's that
do not contain any other point. The arrangement is chiral, and there are
all-together, eight sets of four that do this, given this particular
arrangement.

The complex-polygon equivalent of this is that 3{3}3{3}3{3}3 has
diametric 3-spaces, of the form 3{3}3{3}3{4}2, which four of these cross
only at the centre.

Added 4 February 2019:

My team has had a break-thru involving Wendy's kernel sets of 16 tetrahedra in relation to OEIS sequence A081706 from 0 to 63. If you have been following this thread and are interested to see the nature of this breakthru, send me an email at halitsky.d@att.net ...

• 20 December 2018 - added to the answer to explain why the answer is important to my research team . . . – David Halitsky Dec 20 '18 at 15:08
• 14th version of the answer. Some time ago, a moderator wrote, "I suggest you add any additional news in the form of comments on the accepted answer." Perhaps you missed the word, comments. – Gerry Myerson Feb 4 at 20:54
• @GerryMyerson - thank you!!! I DID misinterpret, and that explains why Todd made the comment he just did on my recent post in Meta. Thank you very much again!! – David Halitsky Feb 4 at 21:01