Does the following linear representation of 112 roots of $E_8$ follow from the known relationship between the (extended) Hamming code and the $E_8$ lattice?

Naively and intuitively, I don't think so. But naivete and intuition only work well during what C.S Peirce called the "abductive" phase of discovery, not afterwards. So please take a look and see what you think.

Edited 12/27/2017 to add: Please see also this follow-up question, which gives the linearization of the remaining 128 roots of $E_8$:

Is there an internally consistent nearest-neighbor relation in this complete linearization of the 240 roots of $E_8$?

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:


12,34  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:


98,76  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:


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.

Edited 12/27 to add: Please also see this related "bounty" question:

Is there an internally consistent nearest-neighbor relation in this complete linearization of the 240 roots of $E_8$?

  • $\begingroup$ David, I will try and shed some light to the connection between $E_8$ and the extended Hamming code as an answer to some of your questions. It just happened that my old laptop froze last Friday, so I could only buy a new one today. It is still a bit lacking in basic software like TeX and Mathematica, so I cannot promise a schedule. Anyway, looks like you found people who know more about polytopes, so that may actually be better! $\endgroup$ – Jyrki Lahtonen Dec 27 '17 at 20:50
  • $\begingroup$ @JyrkiLahtonen - thanks so much for taking the time to respond, Jyrki. But "No !!!!" to your comment about whether the polytope folks will suffice. The polytope folks are essential because they help me (and any other interested parties) visualize certain symmetries of the roots and subsets of roots. But in order to fully understand my "linear" realization of the root-system of $E_8$ (assuming my tuples DO in fact realize this root-system), your insights and experience will, in my opinion, be absolutely crucial. So thanks so much for your patience and willingness to help here. $\endgroup$ – David Halitsky Dec 27 '17 at 21:02
  • $\begingroup$ @JyrkiLahtonen - also, if you haven't yet taken a look at my "bounty" question, I'd be grateful if you'd at least take a look: math.stackexchange.com/questions/2578813/… Even if you aren't interested, you might be able to suggest some profitable approaches to someone who is looking at the problem for me (off-line) - if the question has an affirmative answer, then I think this guy might be able to find it (he dropped out of an MIT math major because the courses were too easy for him . . .!) $\endgroup$ – David Halitsky Dec 27 '17 at 21:19
  • $\begingroup$ I think that your alphabet has nine letters, and the strings have length two. Normally the length of a string tells the number of letters in it. So a string of length nine would be something like $123456789$ or $144255366$. I was a bit confused by your choice of words for a while :-) $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 17:44
  • $\begingroup$ @JyrkiLahtonen - no - as I said - "over ANY alphabet A", which means that the number of letters in the alphabet is irrelevant and immaterial. I am using the numbers 12345679 to denote an ordered n-tuple of 9 letters over A, i.e. the length of the word over A is 9. It probably would have been clearer if I had said abcdefghi instead of 123456789, in which case this (12,23) would be this (ab,bc). (continued next comment) $\endgroup$ – David Halitsky Jan 2 '18 at 17:57

The 240 vertices break into 128 verticies of a half-8-cube, and 112 vertices of the truncated 8-orthotope.

If you extend the packing of the 112 alone, you end up with the 8d semi-cubic, that is, the sets of 8 integers with even sum.

The Hamming-Code is simply the vertices of E8 in terms of the ordinary 8-cubic. You take some code like 0F, and this expands to (0,0,0,0,½,½,½,½). You then replicate this bit sequence for all sixteen coordinates, and you get a cube-cell with 16 vertices of the E8. The Leech-code does the same thing with the Leech-lattice in 24 dimensions.

I found out something very interesting about the Hamming-code: It can be constructed from the representation of E8 as D4D4 * 4. But because D4D4 can be constructed in six different ways, there is no dependency of the second nibble on the first, even if you suppose 0000 0000 and 1111 1111 is part of the code. Without this, there are 24 different Hamming-codes on this construction.

The 112 roots given here belong to an inscribed B8. In order to reach E8, you need to get an even-number of -1 in all positions.

They represent the eutactic star of B8. 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) are adjacent.

My understanding of your scheme is to suppose the nine numbers sit like this:

 1    2    3    4    5    6    7    8    9   +
 9    8    7    6    5    4    3    2    1   -
  +1    +1   0     0    0    0    0    0      (1,2)(2,3)   F   N
  +1    -1   0     0    0    0    0    0      (2,3)(1,2)   F   R
  -1    +1   0     0    0    0    0    0      (9,8)(8,7)   B   N
  -1    -1   0     0    0    0    0    0      (8,9)(7,8)   B   R

This is a trace of the petrie polygon of B8, excluding one of the eight edges. I'm pretty sure that Coxeter has something to say on this, but i would have to look up that book. I think the PP for B8 is 14. This is then in some sort of right product with a group of 8 to give 112.

But i will have to play around with it tomorrow.

  • $\begingroup$ @WendyKrieger - thank you very much. That is a very helpful "encapsulated visualization". I am wondering if you can see a way to apply it in order to answer the follow-up question posed here: math.stackexchange.com/questions/2578813/… $\endgroup$ – David Halitsky Dec 27 '17 at 9:25
  • $\begingroup$ I don't really understand the question. I just visualise these things. $\endgroup$ – wendy.krieger Dec 27 '17 at 9:40
  • $\begingroup$ @wendykrieger - and you do that wondrously well !!! $\endgroup$ – David Halitsky Dec 27 '17 at 9:43
  • $\begingroup$ @DavidHalitsky Updated the answer with many things that i see here. $\endgroup$ – wendy.krieger Dec 29 '17 at 12:18
  • $\begingroup$ @WendyKrieger - yes - your understanding of the scheme is exactly correct. But I'm not sure I understand your comment about the even number of -1's required to get $E_8$. These would NOT be part of the group's root-system, correct? Also, you should really be looking at the situation in terms of my follow-on question, of which this question is Part I. See math.stackexchange.com/questions/2578813/… This second post gives the representation of the 128 as well as the 112. $\endgroup$ – David Halitsky Dec 29 '17 at 12:35

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