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There exists an explicit bijection (due to Cayley, that has built up a very nice table to describe this) between the positive roots of the lattice $E_7$ and $\mathbb{F}_2^6 \setminus \{0\}$ (where $\mathbb{F}_2$ is the field with two elements. Btw, this also preserves orthogonality. There is also a relation between $E_8$ and $\mathbb{F}_2^8 \setminus \{0\}$. Is there an explicit description of the features of this relation in the literature?

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  • $\begingroup$ I don't understand: $E_8$ has 120 positive roots, but $2^8-1=255$. $\endgroup$
    – abx
    Commented Aug 12, 2022 at 13:55
  • $\begingroup$ Yes, you are right, I am quite confused. The reduction modulo 2 of $E_8$ must have a quadratic form with values in $\mathbb{F}_2$ hence there is a non trivial map from $W(E_8)$ to the orthogonal group $O^+(8,\mathbb{F}_2)$ (with kernel $\pm 1$). So what is the relation with $\mathbb{F}_2^8$? $\endgroup$
    – IMeasy
    Commented Aug 12, 2022 at 14:07
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    $\begingroup$ The positive roots correspond to vectors with $q=1$ for the induced quadratic form. This is described for instance in Exercise 1 of Bourbaki Lie groups and Lie algebras, ch. VI, §4. $\endgroup$
    – abx
    Commented Aug 12, 2022 at 14:21
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    $\begingroup$ A construction of the E8 root system is described here: en.wikipedia.org/wiki/E8_(mathematics)#E8_root_system $\endgroup$ Commented Aug 12, 2022 at 17:27
  • $\begingroup$ I have edited the question to make a better link with the answers. $\endgroup$
    – IMeasy
    Commented Aug 13, 2022 at 16:04

1 Answer 1

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Let $L$ denote the root lattice of $E_8$. The group $W(E_8)$ acts (linearly) on $L/2L \cong \mathbb{F}_2^8$, and hence as a permutation of $2^8=256$ elements. This permutation action has the following three orbits:

  1. $\{0\}$, an orbit of size 1.
  2. The classes represented by roots, an orbit of size 120. There are 240 roots in $L$, but each root is equivalent to its negative mod $2L$. In particular, each element of this orbit is represented by a unique positive root.
  3. The frames, an orbit of size 135. The $E_8$ lattice has $240\cdot9$ vectors of length$^2=4$, and each one is congruent to 16 such vectors (including itself). For example, in a rather standard basis (in which $E_8 \supset D_8 \subset \mathbb{Z}^8$), the vector $(2,0^7)$ is congruent to the 16 vectors $(0^a, \pm 2, 0^{7-a})$.

Incidentally, the stabilizer of a root is $W(E_7)$, whereas the stabilizer of a frame is $W(D_8)$. You do indeed find that $\#W(E_8) = 120\cdot \#W(E_7) = 135 \cdot \#W(D_8)$.

For further details, see for example Section 8.1.2 of Berkeley lectures on Lie and quantum groups. (That section is based on lectures by Richard Borcherds, an expert in lattices.)

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