The Weyl group of E8 versus $O_8^+(2)$

Question

Right now Wikipedia says:

The Weyl group of $\mathrm{E}_8$ is of order 696729600, and can be described as $\mathrm{O}^+_8(2)$.

The second part feels wrong to me. $\mathrm{O}^+_8(2)$ is the group of linear transformations of an 8-dimensonal vector space over $\mathbb{F}_2$ preserving a quadratic form of plus type, meaning a nondegenerate quadratic form that vanishes on some 4-dimensional subspaces. The Weyl group of $\mathrm{E}_8$ has center $\mathbb{Z}/2$, consisting of the transformations $\pm 1$. I believe $\mathrm{O}^+_8(2)$ has trivial center. I suspect that $\mathrm{O}^+_8(2)$ is the quotient of the Weyl group by its center.

Is this correct?

Answer 1

Here's the relationship between the $E_8$ Weyl group $W$ and the simple group that might be called $O_8^+(2)$ by some authors, as I understand it. Some of this is contained in or follows from Daniel Allcock's "Ideals in the Integral Octaves" paper from 1998, and work of Conway et. al..

Let $\Omega$ be the $E_8$ lattice, which (up to scaling) I prefer to view as Coxeter's ring of integral octonions, with quadratic form given by the octonion norm $N: \Omega \rightarrow {\mathbb Z}$. Then $W$ acts on $\Omega$ by norm-preserving ${\mathbb Z}$-linear automorphisms. Reducing mod two, $W$ acts by ${\mathbb F}_2$-linear automorphisms on $\bar \Omega = \Omega / 2 \Omega$, preserving the reduction of the norm $\bar N : \bar \Omega \rightarrow {\mathbb F}_2$ (a quadratic form mod $2$).

This gives a homomorphism from $W$ to the orthogonal group $O(\bar \Omega, N)$, but not a homomorphism from $W$ to the special orthogonal group $SO(\bar \Omega, N)$; this is because the special orthogonal group is defined via the Dickson map rather than the determinant in characteristic two. The group $SO(\bar \Omega, N)$ is a finite simple group which might be called $O_8^+(2)$ by some authors. Related to Noam's comment, the subgroup $SO(\bar \Omega, N)$ can be defined as the subgroup of the orthogonal group $O(\Omega, N)$ consisting of elements fixing an even-dimensional subspace (one can realize the Dickson invariant here as the dimension of the fixed space, mod $2$). A simple Weyl reflection will fix a 7-dimensional subspace and 7 is odd :)

The moral is that there's not an interesting homomorphism from $W$ to $SO(\bar \Omega, N)$.

But... if one takes the even subgroup $W^+ \subset W$, which coincides with the commutator subgroup $[W,W]$, then the image of $W^+$ in $O(\bar \Omega, N)$ coincides with $SO(\bar \Omega, N)$. The kernel of this homomorphism is the central subgroup $\{ \pm 1 \}$ in $W^+$. Thus there's a central extension, $$1 \rightarrow \{ \pm 1 \} \rightarrow W^+ \rightarrow SO(\bar \Omega, N) \rightarrow 1.$$

So to summarize, the Weyl group $W$ contains $W^+$ with index two, and the simple group $SO(\bar \Omega, N)$ is a quotient of $W^+$ by a central subgroup of order two.

Answer 2

Let me describe this in my language. $E_8$ lattice has 120 axes (240 vectors). Weyl group of $E_8$ Lie group (let's call it $W(E_8)$) is generated by reflections in these vectors. This is conjugacy class of size 120 in $W(E_8)$. It contains elements of determinant $-1$. If we take just "rotations" in Weyl group i.e. elements of determinant $1$ then we obtain group of size 348 364 800 which is $2.O_8^+(2)$ according to Atlas notation.

The center of $W(E_8)$ contains two elements $\pm I$. The simple group $O_8^+(2)$ of size 174 182 400 is obtained by dividing "rotation" part of $W(E_8)$ by two elements center.

The proof that obtained factor group is really $O_8^+(2)$ may go via octonions over field $\mathbb F_2$. Taking $E_8$ lattice modulo 2 (which is not easy for my intuition) we obtain 120 vectors of length 1 and 135 vectors of length $\sqrt 2$ (sums of perpendicular pairs of length 1). Multiplication of integral octonions is then mapped to multiplication in octonions over $\mathbb F_2$ (let's call it $\mathbb O_{F_{2}}$). There are 120 invertible elements and 135 zero divisors in $\mathbb O_{F_{2}}$. Each $W(E_8)$ element is mapped to norm preserving automorphism i.e. it permutes 120 points of norm 1 and 135 points of norm 0. In order to obtain representation of this group as matrices 8x8 over $F_2$ we need to fix some basis in 255 points of $\mathbb O_{F_{2}}$. Actually it will be $O_8^+(2).2$, because $-I$ does not change points of $\mathbb O_{F_{2}}$.

The size of the group can be calculated as 240*126*60*2 * 24*2*2*2 = 696 729 600, because for standard basis $e_0,...,e_7$ first vector $e_0$ can be mapped to any of the 240 vectors. The second vector $e_1$ can be mapped any of the 126 perpendicular. Third one $e_2$ can be mapped any of the 60 vectors perpendicular to given perpendicular pair of vectors. The axis of fourth vector is already defined - for given perpendicular triple there is just one fourth vector in $E_8$ lattice such that resulting 4-space contain $D_4$ sublattice of $E_8$. We may also see that if first vector is octonion $1$ then fourth vector is $\pm$ product of second and third one. Remaining four vectors are determined by the fact that they belong to perpendicular $D_4$ lattice to the one generated by first four vectors - there are just 24 vectors there to use.

I see now that wikipedia article about $E_8$ is corrected.