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 http://categorified.net/LieQuantumGroups.pdf. (That section is based on lectures by Richard Borcherds, an expert in lattices.)

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.)

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 $240\cdot9$ roots 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 http://categorified.net/LieQuantumGroups.pdf. (That section is based on lectures by Richard Borcherds, an expert in lattices.)

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 http://categorified.net/LieQuantumGroups.pdf. (That section is based on lectures by Richard Borcherds, an expert in lattices.)