Any Existing Time-Division Multiplexing Algorithm in which Slot-Allocation is Governed by the $E_8$ Root System?

Question

Background:

https://en.wikipedia.org/wiki/Time-division_multiplexing

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

Question:

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.

Answer 1

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

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.

Answer 2

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.