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Let $\{1, \omega, \omega^2\}$ be the three cube-roots of one.

Define the Eisenstein integers, $\Bbb{E}$, to be the $\Bbb{Z}$-linear combinations of $1$ and $\omega$. Note that $\omega^2 = -1-\omega \in \Bbb{E}$.

Let $\lambda = 1-\omega \in \Bbb{E}$. If we identify elements of $\Bbb{E}$ that differ by a multiple of $\lambda$, we obtain three equivalence classes, with representative elements $\{0, 1, -1\}$. Let $c: \Bbb{E} \to \{0,+, -\}$ be the classifier function that tells us which equivalence class a given integer belongs to.

Define the Tetracode, $T$, to be the following set (which is a perfect linear error-correcting code with Hamming distance 3):

$$\left\{\begin{matrix} (0,0,0,0), & (0,+,+,+), & (0,-,-,-), \\ (+,0,+,-), & (+,+,-,0), & (+,-,0,+), \\ (-,0,-,+), & (-,+,0,-), & (-,-,+,0) \end{matrix}\right\}$$

Let $E_8' = \left\{(w, x, y, z) \in \Bbb{E}^4 \mid (c(w), c(x), c(y), c(z)) \in T\right\}$. It is an 8-dimensional lattice, in which every point has 240 nearest neighbours. (24 of those neighbours are found by changing one coordinate, and 216 are found by changing three coordinates).

Is $E_8'$ equivalent to $E_8$?

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Short answer: yes, because $E_8$ is the unique eight-dimensional lattice with kissing number 240 (the largest possible). (Reference: Theorem 8, Chapter 14 by Bannai and Sloane, from Conway and Sloane, Sphere packings, lattices and groups, 1988.)

(Edit to add: your construction is an instance of what Sloane calls "Construction A for complex lattices", from Section 8, Chapter 7 of Conway and Sloane, whereby you can combine a length $n$ $q$-ary code with an index $q$ ideal of $\Bbb{E}$ or similar to form a complex $n$-dimensional lattice. Example 11b notes that the Tetracode yields $E_8$.)

If you want an explicit isomorphism, then write elements of $E_8'$ as $v=(a,b,c,d,e,f,g,h)\in\Bbb{Z}^8$, representing $(a+b\omega,c+d\omega,e+f\omega,g+h\omega)\in\Bbb{E}^4$. The triality vector $(c(a+b\omega),c(c+d\omega),c(e+f\omega),c(g+h\omega))$ is then $(a+b,c+d,e+f,g+h)\in T$, taking the co-ordinates modulo 3. Define

$$ M=\begin{pmatrix} \begin{array}{rrrrrrrr} 0 & 0 & 0 & 0 & 3 & 0 & 3 & 0 \\ 0 & 0 & -2 & -2 & 1 & -2 & 1 & -2 \\ 0 & 0 & -2 & 4 & 1 & -2 & 1 & -2 \\ 0 & 0 & 4 & -2 & 1 & -2 & 1 & -2 \\ 0 & 0 & 0 & 0 & 3 & 0 & -3 & 0 \\ -2 & -2 & 0 & 0 & 1 & -2 & -1 & 2 \\ -2 & 4 & 0 & 0 & 1 & -2 & -1 & 2 \\ 4 & -2 & 0 & 0 & 1 & -2 & -1 & 2 \\ \end{array} \end{pmatrix} $$

Then, considering an element of $E_8'$ as a column vector in $\Bbb{Z}^8$ as described above, multiplying on the left by $\frac16 M$ will give an element of the standard version of $E_8$:

$$\Gamma_8 = \left\{(x_i) \in \mathbb Z^8 \cup (\mathbb Z + \tfrac{1}{2})^8 : {\textstyle\sum_i} x_i \equiv 0\!\!\pmod 2\right\}.$$

Of course this is one of many possible maps since $E_8$ has a large symmetry group, but I don't think there is a much nicer form of the matrix. Maybe it is not so surprising that $M$ will look a bit ragged, since the definition of the Tetracode used a particular choice of (non-symmetric) basis and $M$ has to depend on this choice.

To check this works, we first need to see that $x=\frac16 Mv\in E_8$ for $v\in E_8'$. First check that each component of $Mv$ is congruent to zero mod 3. Note that pairs of entries in $M$ satisfy $M_{2i,j}\equiv M_{2i+1,j}\pmod 3$, which means that $$6x_i\equiv M_{i,0}.c(a+b\omega)+M_{i,2}.c(c+d\omega)+M_{i,4}.c(e+f\omega)+M_{i,6}.c(g+h\omega)\pmod 3.$$ Modulo 3, the only values of $(M_{i,0},M_{i,2},M_{i,4},M_{i,6})$ that arise are $(0,0,0,0)$, $(0,1,1,1)$, and $(1,0,1,2)$. Dotting these with a Tetracode vector, $(c(a+b\omega),c(c+d\omega),c(e+f\omega),c(g+h\omega))$, gives zero modulo 3 (easily checked; also follows from the fact that the Tetracode is self-dual). Thus $6x_i\equiv0\pmod3$, and so $x_i\in\frac12\Bbb{Z}$.

To see $x_i\equiv x_j\pmod1$, note that each column of $M$ is constant modulo 2. Thus $6x_i\equiv 6x_j\pmod2$, which is what we want since $6x_i$ and $6x_j$ are also zero modulo 3.

To complete the proof that $x\in E_8$, add the rows of $M$ to see that $$\sum_i x_i=\frac16 \sum_{ij} M_{ij}v_j=2(e-f)\equiv0\pmod2.$$

Finally, to show that the image $\frac16 M(E_8')$ is the whole of $E_8$, it's sufficient to check that we can reach a basis of $E_8$.

$$\frac16 M\begin{pmatrix} \begin{array}{rrrrrrrr} 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & 0 \\ -1 & -1 & 1 & 1 & -1 & 0 & 0 & 0 \\ -1 & -1 & 2 & -1 & 0 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 & 1 & -1 & 0 & 1 \\ 0 & -1 & 0 & 0 & 1 & -1 & 0 & -1 \\ 1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 \\ \end{array} \end{pmatrix}= \begin{pmatrix} \begin{array}{rrrrrrrr} 1 & -1 & 0 & 0 & 0 & 0 & 0 & 1/2 \\ 1 & 1 & -1 & 0 & 0 & 0 & 0 & 1/2 \\ 0 & 0 & 1 & -1 & 0 & 0 & 0 & 1/2 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1/2 \\ 0 & 0 & 0 & 0 & 1 & -1 & 0 & 1/2 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 & 1/2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1/2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1/2 \\ \end{array} \end{pmatrix} $$

The columns of the matrix on the left are all in $E_8'$, having triality vectors $(0,1,1,1)$, $(0,1,1,1)$, $(0,0,0,0)$, $(0,0,0,0)$, $(0,2,2,2)$, $(1,0,1,2)$, $(0,0,0,0)$, $(0,0,0,0)$ respectively. The columns of the matrix on the right form a basis of $E_8$.

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  • $\begingroup$ Wow, you've done a lot of work! Thank you very much. $\endgroup$ – apt1002 Feb 15 '17 at 16:13

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