Does anyone know of any existing Time-Division Multiplexing algorithm in which competing signals are allocated slots in the physical channel in a particular way defined by the internal structure of the root-system of $E_8$ (when this root-system is expressed in terms of a basis which mathematicians customarily use?)

Example (trivial):

In one familiar expression of the root-system of $E_8$ in 8-space, the 240 roots wind up in two natural subsets of 128 and 112.

Since 128:112 = 8:7, the Time-Division Multiplexing algorithm would allocate physical channel slots to two competing signals in the ratio of 8 slots (for the more important signal) to 7 slots (for the less important signal).

Thanks for whatever time you can afford to spend considering this question.

  • $\begingroup$ @TobiasKildetoft $\endgroup$ – David Halitsky Dec 11 '17 at 10:36
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    $\begingroup$ It was not time-division, but I once tested the use of $E_8$-modulation for European digital tv broadcast (a former graduate student of mine was heavily involved in the study group, and I wanted to give her a hand). Not all of it is relevant to your project, but "the slides with color in them" try to explain how we can build $E_8$ within a fourfold Cartesian power of the square lattice by introducing "color code constraints". I put my PowerPoint slides up - just in case. I'm sad to say this was not included to the DVB standard. $\endgroup$ – Jyrki Lahtonen Dec 11 '17 at 10:42
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    $\begingroup$ FYI: Within a single thread you cannot @-ping users who have not participated in that thread. So Tobias will not see your ping from here. $\endgroup$ – Jyrki Lahtonen Dec 11 '17 at 10:43
  • $\begingroup$ @JyrkiLahtonen !!!!!!!!!!! !!!!!!!!!!!!!!! !!!!!!!!!!!!!!! That's really quite remarkable ! Thanks so much for the link! In the case of DNA and genes, it's pretty obvious (once you think about it) why genes MUST be TDM signals . . . $\endgroup$ – David Halitsky Dec 11 '17 at 10:44
  • $\begingroup$ @JyrkiLahtonen - I just looked at your slides! Great job! Also, you're going to laugh when you see our decomposition and how it relates to energetics (more precisely - to the relative "delta-H" enthalpies of Watson-Crick base pairing across polynucleotide segments in double-helix conformation. $\endgroup$ – David Halitsky Dec 11 '17 at 14:25

I recap how I described $E_8$ in that slide set, and how the so called "Construction A" associates $E_8$ and the extended Hamming code.

Consider the integer lattice $\Lambda_0=\Bbb{Z}^2$. We partition it into four cosets of the sublattice $\Lambda=2\Lambda_0$. I used colors for the cosets as follows

planar integer lattice partitioned into four cosets of the doubled sublattice

In other words "Red" = $\Lambda$, "Blue"=$(1,0)+\Lambda$, "Black"=$(0,1)+\Lambda$, "Green" =$(1,1)+\Lambda$.

The lattice $E_8$ is easy to describe as a subset of the 4-fold Cartesian power $\Lambda_0^4$. We include the vectors $(x_1,x_2,x_3,x_4)\in\Lambda_0^4$ such that:

  • All the four components $x_i$ have the same color.
  • Two are red and two are green, in any order (so for example GRRG and GGRR are ok).
  • Two are blue and two are black, in any order.

This gives us the following census for the 240 roots (recall that a zero-component is red), they are the vectors with length two (or squared length four):

  • A single red component of length two and three zero components. There are four available non-zero red components, and it can be placed into any of the four components. A total of $4\cdot4=16$ vectors.
  • Two green components of length $\sqrt2$ and two zero components. There are four available green components, and there locations can be selected in $\binom 42=6$ different ways so we get $4\cdot4\cdot6=96$ vectors like this.
  • There are two blue components of length one, so we get $2^4=16$ all-blue vectors. Similarly we get $16$ all black vectors.
  • The number of blue and black mixtures is $2^4\cdot6=96$ because we can choose the black position in six different ways, and there are two choices for each component.
  • $16+96+16+16+96=240$.

Using Construction A we would need the $16$ words of the Hamming code. Those are $0000000$, $11111111$, $00001111$, $11110000$, $00110011$, $00111100$, $11000011$, $11001100$, $01010101$, $01011010$, $01100110$, $01101001$, $10010110$, $10011001$, $10100101$ and $10101010$. The lattice $E_8$ then consists of the vectors in $\Bbb{Z}^8$ that are congruent to one of those sixteen vectors modulo two. Observe that:

  • We can permute the 8 components at leisure to arrive at different but isometric copies of $E_8$. Exactly $168$ out of the $8!=40320$ permutations give rise to the same version of $E_8$. The shown bit ordering matches with my "color code". For example the vectors congruent to $11000011$ have colors GRRG.
  • When looked at this way the 240 vectors in the root system $E_8$ are the sixteen vectors with single component $\pm2$ together with seven zeros, and the $16\cdot14=224$ vectors with $\pm1$ at a set of four positions matching with the positions of $1$s in some word of the Hamming code and zeros elsewhere.
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    $\begingroup$ I don't see a connection to time division, so I make this CW. Just in case the link to my slides rots some time in the distant future. $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 17:26
  • $\begingroup$ thanks very much for taking the time to do that ! $\endgroup$ – David Halitsky Jan 2 '18 at 18:15

The comment by @JyrkiLahtonen above is actually a wonderful answer to the question, even though he didn't post his comment as an answer because his use of $E_8$ did not directly involve TDM. But even though his work did not involve TDM, he was kind enough to post some slides which immediately told me how and why his work is obviously related to the work that my team is currently engaged in, so I am "declaring" his comment as an official answer.

  • $\begingroup$ @JyrkiLahtonen - I just made you comment an "official" answer - hop you don't mind . . . $\endgroup$ – David Halitsky Dec 11 '17 at 14:40
  • $\begingroup$ Similarly to how I was not notified when you used @myname as a comment on the question, Jyrki will not be notified of this one, since he has only commented on the question, not this answer. $\endgroup$ – Tobias Kildetoft Dec 11 '17 at 15:14
  • $\begingroup$ @TobiasKildetoft - thanks for letting me know Tobias - hope you are well, and thanks again for the diagram ! $\endgroup$ – David Halitsky Dec 11 '17 at 15:16
  • $\begingroup$ @TobiasKildetoft - if and when you have a chance, please see this question at MO (a follow-up to my MOquestion earlier today) mathoverflow.net/questions/288394/… $\endgroup$ – David Halitsky Dec 13 '17 at 1:37
  • $\begingroup$ @TobiasKildetoft - please see this reworded question at MO - I reworded the question as per suggestion from mods $\endgroup$ – David Halitsky Dec 14 '17 at 15:44

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