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Edited 1/21/2018 to add the following:

Here is a DropBox link

https://www.dropbox.com/s/7rtt0iqmgimsgzu/Zumkeller_edge-magic.pdf?dl=0

to a PDF showing how my team used biomolecular first principles to arrive at a set of 240 biomolecular objects (which we believe to be an instantiation of the roots of $E_8$), and more generally, how we arrived at related sets of biomolecular objects with the cardinalities of the Zumkeller numbers (176,240,336) and the correponding edge-magic injection label numbers (11,15,21). The set of 240 objects defined in the attached PDF can be related, in various ways, to the linearization of the 240 roots of $E_8$ described below in the original question.

Original statement of question

Below is a complete linearization of the 240 roots of $E_8$.

Part I, which is a linearization of 112 roots, was already given in this recent question:

Does this linear representation of 112 roots of $E_8$ follow from the (extended) Hamming code's relation to the $E_8$ lattice?

Part II is a linearizaiton of the remaining 128 roots.

Question:

Is there an internally consistent nearest-neighbor relation in the complete linearization enumerated below of all 240 roots?


[Edited to add:} By "internally consistent", I mean a relation defined on the 240 tuples given below such that according to this relation, EACH tuple has 56 nearest neighbors, 126 "second" nearest-neighbors , 56 "third" nearest-neighbors, and finally, 1 "antipodal" (opposed) tuple "beyond" these 56 "third" nearest neighbors. See the section entitled "Coordinates" here:

https://en.wikipedia.org/wiki/4_21_polytope

and in particular, the third paragraph of this section beginning: "Each vertex has 56 nearest neighbors ..."


Enumeration of linearized roots:

Part I: Linear representation of 112 roots of $E_8$:

Given any ordered string S of 9 letters over any alphabet A , S obviously contains only 8 ordered 2-tuples composed of adjacent letters, and therefore:

1) only 28 possible choices of UNordered pairs of such 2-tuples:

12,23
12,34  23,34
12,45  23,45  34,45
12,56  23,56  34,56  45,56
12,67  23,67  34,67  45,67  56,67
12,78  23,78  34,78  45,78  56,78  67,78
12,89  23,89  34,89  45,89  56,89  67,89  78,89

2) only 56 possible choices of ORDERED pairs of such 2-tuples, e.g. (23,12) as well as (12,23), etc.

Now suppose that we are allowed to read S "forward" or "backward". Then corrsponding to (1) and (2), we will have:

3) only 28 possible choices of UNordered pairs of 2-tuples:

98,87
98,76  87,76
98,65  87,65  76,65
98,54  87,54  76,54  65,54
98,43  87,43  76,43  65,43  54,43
98,32  87,32  76,32  65,32  54,32  43,32
98,21  87,21  76,21  65,21  54,21  43,21  32,21

4) only 56 possible choices of ORDERED pairs of such 2-tuples, e.g. (87,98) as well as (98,87), etc.

Hence, we will have 112 ordered pairs of ordered 2-tuples, each unqiuely identified by:

5) an index R indicating which way we R(ead) a 9-tuple (backward or forward)

6) an index O indicating which way we chose to O(rder) each Unordered pair of 2-tuples (in its "natural" order relative to our initial choice of read-direction, or in its "reverse" order relative to our initial choice of read-direction.)

7) an index L indicating which of the 28 pairing L(ocations) we chose

And clearly:

8) the 112 resulting (R,O,L) triples can be mapped in an obvious way onto the 112 roots of $E_8$ which have integer entries when the roots of $E_8$ are coordinatized in the usual way:

https://en.wikipedia.org/wiki/E8_(mathematics)

For example

        Root              L         R          O
(+1,+1,0,0,0,0,0,0)    (12,23)   forward    natural
(+1,-1,0,0,0,0,0,0)    (23,12)   forward    reverse
(-1,+1,0,0,0,0,0,0)    (98,87)   backward   natural
(-1,-1,0,0,0,0,0,0)    (87,98)   backward   reverse

9) the 112 (R,O,L) triples can be paired off in an obvious way as inverses of one another.

Part II: Linearization of remaining 128 roots of $E_8$

10) exactly 28 ORDERED 2-tuples of NON-adjacent letters when the string S is read FORWARD

13
14 24
15 25 35
16 26 36 46
17 27 37 47 57
18 28 38 48 58 68
19 29 39 49 59 69 79

11) exactly 28 ORDERED 2-tuples of NON-adjacent letters when the string S is read BACKWARD

31
41 42
51 52 53
61 62 63 64
71 72 73 74 75
81 82 83 84 85 86
91 92 93 94 95 96 97

(Note that these are naturally definable as inverses of the 28 in (10) 

12) exactly 35 ORDERED 3-tuples of NON-adjncent letters when the string S is read FORWARD:

135
136 146
137 147 157
138 148 158 168
139 149 159 169 179      
    246  
    247 257
    248 258 268  
    249 259 269 279      
        357
        358 368
        359 369 379       
            468
            469 479       
                579       

13) exactly 35 ORDERED 3-tuples of NON-adjacent letters when the string S is read BACKWARD:

531
631 641
731 741 751
831 841 851 861
031 941 951 961 971      
    642  
    742 752
    842 852 862  
    942 952 962 972      
        753
        853 863
        953 963 973       
            864
            964 974       
                975

 (Note that these are naturally definable as inverses of the 35 in (12) 

14) exactly 1 ordered 5-tuple of NON-adjacent letters when the string S is read FORWARD

13579 

15) exactly 1 ordered 5-tuple of NON-adjacent letters when the string S is read BACKWARD:

97351 (inverse of the 5-tuple in (14)

Edited 12/29/2017

These definitions of 1st/2nd/3rd nearest neighbors among the 112 root coordinate 8-tuples in 8-space were provided by Wendy Krieger in response to my question which preceded this one:


Two points are close neighbours, if they share one coordinate of the same sign and position, eg (2,2,0,0,0,0,0,0) and (2,0,0,0,0,-2,0,0) "

They are second neighbours, if they share no coordinate, or one opposite sign, eg (2,2,0,0,0,0,0,0) is at right angles to (2,-2,0,0,0,0,0,0) and to (0,0,2,0,0,0,2,0,0).

The third-order is if they share a coordinate, with opposite sign, (-2,0,2,0,0,0,0,0), and opposite if both signs are reversed.


Edited 12/30 to add the following information:

Note that of the 240 given 2-tuples, only the 112 2-tuples of adjacent letters can be specified energetically. This is because over the DNA {tcag} alphabet or the RNA {ucag} alphabet, these 2-tuples have associated "relative-delta-H enthalpies" as follows. (These indicate relative strength of complementary binding of these 2-tuples across the two strands in duplex ("double-helix") DNA or RNA - the values below are for RNA, not DNA).

aa  2.80
ga  1.41
ua  2.07
ca  1.16
ag  1.52
au  2.86
gg  0.27
ac  1.91
ug  1.16
cg  0.00
gu  1.91
gc  0.95
uu  2.80
cu  1.52
uc  1.41
cc  0.27
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  • $\begingroup$ @user296602 - I have edited the post to clarify what I mean by "internally consistent" nearest-neighbor relation - I hope you and the other voters will find this edit satisfactory, and thanks as always for your patience $\endgroup$ – David Halitsky Dec 26 '17 at 5:28
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    $\begingroup$ It is evident in the sources you linked to that the $240$ $E_8$ roots form a uniform polytope on the $7$-sphere (embedded in dimension $8$) and that the vertices have coordinates that are particularly convenient for the distinction between subsets of $112$ and $128$ roots. So it is straightforward to compute the distances between pairs of roots in this representation. Does your problem consist of verifying the "nearest neigbhor" edges of the polytope conform to your "internally consistent" counts of first-, second-, and third-degree vertex adjacency? $\endgroup$ – hardmath Dec 28 '17 at 18:50
  • $\begingroup$ @hardmath - thanks for taking the time to think about this question - I am very grateful. As I see it, our problem can be stated this way. We want to first define a nearest-neighbor relation on our 240 given tuples (making NO reference to $E_8$ roots at all), and then to show that there is an isomorphism from our 240 tuples to the standard coordinate 8-tuples of $E_8$ roots such that this isomorphism is STRUCTURE-PRESERVING in the sense that it respectively maps 1st, 2nd, and 3rd nearest neighbors in our set onto 1st, 2nd, and 3rd nearest neighbors in the root-system of $E_8$. $\endgroup$ – David Halitsky Dec 28 '17 at 19:04
  • $\begingroup$ @hardmath - in other words, my team does not yet have a satisfactory definition of nearest-neighbor on OUR tuples which will be preserved under the isomorphism - this definition is what we are looking for. $\endgroup$ – David Halitsky Dec 28 '17 at 19:06
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If you are using the numbers 1-9, in various swapping, i suspect that you are looking at something like A8, which is a subgroup of E8. This is 72+84+84, being the runcinated 8-simplex (vertex figure of A8), and the second rectated 8-simplex and its central invert.

In terms of the general A-coordinates (ie the plane x0+x1+...+x9=0),

The 72 represent the seventy-two pairs +1,-1.

The 84 points are (1,1,1,0,0,0,0,0,0)-1/3, and (1,1,1,1,1,1,0,0,0)-2/3, that is, subtract the fraction from each coordinate. All permutations, no change of sign.

The vertices of the various polytopes in En can be derived by drawing spheres of given sizes on the lattice, and it is the lattices that tell us a lot of what's going on.

The projection via A8

E8 can be constructed from the union of three polytopes of simplex symmetry: /7/, 2/5, and 5/2. This notation uses '/' for a marked node, and a numeral for a chain of unmarked branches. Nodes buffer the ends, so

 /7/  means x---o---o---o---o---o---o---x
 2/5  means o---o---x---o---o---o---o---o
 5/2  means o---o---o---o---o---x---o---o
                                b   a
                             /---------\
 /6B  means o---o---o---o---o---o---o   o   aka 4_21

This notation is part of my extension to Coxeter's 2_21 style, designed to be lining, and free from subscripts or superscripts, so it can acquire these if needed. Straight numbers are a string of unmarked 3 branches. A,B,C jump over 1,2,3 branches at the end, E,G jump over branches at the beginning, Q is a branch marked '4', and F is a branch marked '5'. H is a branch marked '6'.

Nodes are marked with mirror-edges (the usual mark) with /, the dual has mirror-margins (ie faces reflect in their walls), with . So the dual of /4B is \4B.

The vertices of the simplex-symmetry figures, of the style p/q, can be constructed as (q+1) repeated p times, followed by -(p+1) q times. This can then be divided throughout by (p+q+2). When several nodes are marked, evaluate each node separately, and add them together.

In the present case, we would get these, with all permutations.

/7/  gives  8  -1  -1  -1  -1  -1  -1  -1  -1
         +  1   1   1   1   1   1   1   1  -8
     or     9   0   0   0   0   0   0   0  -9    72 vertices

2/5   gives 6   6   6  -3  -3  -3  -3  -3  -3    84 vertices
5/2   gives 3   3   3   3   3   3  -6  -6  -6    84 vertices

We divide through by 3 here, to get the coordinates as before.

The connection with the 9d presentation of 2_21, is that the 6d lattice is simply a grouping of the 8d lattice, with the sum-zero in all cases.

  /7/  gives  (9,-9,0)  (0,0,0)   (0,0,0)      18 vertices = 3 hexagons
  2/5  gives  (6,-3,-3) (6,-3,-3) (6,-3,-3)    27 vertices = tri-triangular prism
  5/2  gives  (3,3,-6)  (3,3,-6)  (3,3,-6)     27 vertices = ditto.

Because these are real points in 9 dimensions, the standard cubic tricks work here. The dot product of two coordinates gives the angle between the position vectors, and hence what ring they fall on. This is why here it's better to divide by 3 throughout. 2 2 2 -1 -1 -1 -1 -1 -1 -1 18 is two copies of the same vector 2 2 2 -1 -1 -1 -1 -1 -1 -1 18 9 gives close neighbours 3 0 0 -3 0 0 0 0 0 0 9 0 gives right or middle neibours 2 -2 -1 2 -1 -1 2 -1 -1 -1 0 -9 gives far neighbours -3 0 0 3 0 0 0 0 0 0 -9 -18 gives opposites -2 -2 -2 1 1 1 1 1 1 1 -18

So the presentation for E6 in 9 dimensions, with three sets of brackets, is a subset of E8 in 9 dimensions, without sets of brackets, in 9 dimensions. Inside each set of brackets, the sum must be zero.

The presentation of E8 in B8

You can make an eight-dimensional representation of E8 from the semi-cubic. Here we use an edge of two-sqrt(2). The standard coordinates for a semi-cubic is a set of even numbers, whose sum is a multiple of four, so eg (2,2,0,0,0,0,0...). A quarter-cubic is made by adding a semi-cubic to the lattice in the centre of the cubes, so we get a set of eight odd numbers that make a sum of four, viz (1,1,1,1,1,1,1,1,1).

The root lattice here is a sphere of radius sqrt(8), so only two sets of vertices count:

          APEC = all permutations, even changes of sign

 1/5A  2,2,0,0,0,0,0,0    112  rectified 8-orthoplex
 6/A   1,1,1,1,1,1,1,1    128  8-halfcube.

We can derive the root lattices for E7 and E6, by replacing the last 2 or 3 coordinates by sqrt(2) [ie q], or sqrt(3) [ie h]. This means that we reduce the eight dimensions by replacing a square or cube by its diagonal.

So the root lattice for 6d is APEC = all perm even change of sign

     2,2,0,0,0, 0           40   truncated 5-orthoplex
     1,1,1,1,1, sqrt(3)     32   elongated half-6cube        

The sphere is sqrt(8) still, so we get an elongated half-cube, and we get the remaining vertices from the vertex-figure of B5.

For the 7d case we can have 2,2 as the equal coordinates merged to 2sqrt2. so we get APEC

    2,2,0,0,0,0   0        60    truncated 6-orthplex
    1,1,1,1,1,1  1+1       64    elongated half 7-cube
    0,0,0,0,0,0  2+2        2    points at the poles.

These vertices are also ordinary cubic coordinates, and so the regular product applies.

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  • $\begingroup$ @WendyKrieger - I have not accepted your answer "officially" because it is not an answer to the question per se. If you can give a nearest-neighbor relation for the linear tuples related to $A_8$ instead of $E_8$, that would be wonderful, and of course, it would qualify as at least one "official" alternative answer. Also, I am grateful to your response here because on the purely empirical side of my team's research, subsets of {84} have appeared in various contexts, and we've been wondering why ! $\endgroup$ – David Halitsky Dec 27 '17 at 9:42
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    $\begingroup$ @DavidHalitsky Looking on how you construct your set, they are restricted to B8 and A8. That is, you should be looking for a body-centering. The vertices of 128 opposite 112, come from 8 half-integers with even sum, and the two sets of 84 in A8, come from adding 1/3 and 2/3 to each of the 9 axies of the plane in question. So i suspect there is an additional shift you have not counted for, if the subgroups are not appearing in your operators. $\endgroup$ – wendy.krieger Dec 27 '17 at 9:55
  • $\begingroup$ @wendykrieger - thanks again for your additional comment here. The reason why we want to restrict attention to $E_8$ at the moment is because $E_8$ has been known for years to be intimately associated with energetics (qua mathematical physics) and in our biomolecular case, the 240 tuples are also intimately associated with the energetics of dinucleotide Watson-Crick pairing with complementary dinucleotides. But if we fall to "get" $E_8$ out of these 240, we will certainly take you suggestion to look at $B_8$ and $A_8$ . . . $\endgroup$ – David Halitsky Dec 27 '17 at 10:02
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    $\begingroup$ The reason that i was suggesting looking for a centering, is because your constructions are limited to $B_8$ anyway, or $A_8$. In order to do E8 in integers alone, you would need to consider sets of even and sets of odd numbers that make a multiple of 4. $\endgroup$ – wendy.krieger Dec 27 '17 at 10:15
  • $\begingroup$ @wendykrieger - I think you may have misinterpreted the way in which integers are being used in the 240 tuples which I gave. These have no meaning as integers per se - they are just "location indicators" - I could as easily have used abcdefghi as 123456789. So, it's still my personal guess that the symmetries in the given 240 are in fact symmetries of $E_8$. but it may well turn out that your proposal is correct, and that we should turn our attention from $E_8$ to $A_8$ and $B_8$ . . . $\endgroup$ – David Halitsky Dec 27 '17 at 10:21
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Sadly I have to correct Wendy in her answer.

The E8 polytope 4_21 was given correctly as the hull of the compound of the hemiocteract x3o3o *b3o3o3o3o3o and the rectified 8-crosspolytope o3o3o *b3o3o3o3x3o. Alternatively 4_21 thus can be considered as zero-height lace tower of these two 8D polytopes.

But the E7 and E6 polytopes had gone slightly wrong. Here comes the corrected version each:

The E7 polytope 2_31 can be given as an axial stack (or lace tower) of 5 vertex layers. These are the vertex sets each of a single point o3o3o *b3o3o3o, a hemihexeract x3o3o *b3o3o3o, the rectified 6D crosspolytope o3o3o *b3o3x3o, again the same hemihexeract x3o3o *b3o3o3o, and finally again an opposite point o3o3o *b3o3o3o. The stacking height between all these layers here is 1/2 of the edge size of all these layer polytopes. Cf. https://bendwavy.org/klitzing/incmats/laq.htm

It is well-known that a lace prism of 2 hemihypercubes of alternate orientations would provide a hemihypercube of the next dimension. But here those hemihypercubes are aligned in the same orientation. In fact these 2 layers truely provide a true unit sized hemihypercubic prism, and those prism edges do contribute to the overall hull! And also the equatorial layer is, as mentioned above, a rectified crosspolytope, not, as Wendy states, a truncated one.

The E6 polytope 1_22 can be provided as hull of 3 vertex layers, the individuals being the vertex set of a hemipenteract x3o3o *b3o3o, a rectified 5D crosspolytope o3o3o *b3x3o and that of the alternate hemipenteract o3o3x *b3o3o. The layer spacing here is sqrt(3/8) = 0.612372. Cf. https://bendwavy.org/klitzing/incmats/mo.htm

This time we truely have alternate 5D hemihypercubes. Thus a somewhat stretched 6D hemihypercube would be correct indeed to describe both extremal layers. But the equatorial layer again is rather a rectified crosspolytope, not a truncated one.

--- rk

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For any of the 112 tuples involving one or two 2-tuples of ADJACENT letters, there is a known precise energetic value which can be calculated for the tuple, using the methodology presented in these two papers:

https://www.ncbi.nlm.nih.gov/pubmed/6616016

http://onlinelibrary.wiley.com/doi/10.1002/bip.360220812/abstract

But for any of the 128 tuples involving NO 2-tuple of adjacent letters, one can only calculate an average energetic value over a set of possibilities represented by the tuple.

My team currently suspects that the desired nearest-neighbor relation may be definable in terms of these precise and averaged energetic values for the complete set of 240 tuples.

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