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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?

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    $\begingroup$ Things are also complicated by the fact that the simple group $O_{8}^{+}(2)$ has a Klein $4$-group as its Schur multiplier. $\endgroup$ Feb 4, 2016 at 1:11
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    $\begingroup$ The Atlas gives the order of $O_{8}^{+}(2)$ as 174182400. $\endgroup$ Feb 4, 2016 at 1:14
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    $\begingroup$ If I remember right the index-2 normal subgroup of $O_{2k}^\pm(2)$ consists of the transformations that have an even-dimensional fixed subspace. $\endgroup$ Feb 4, 2016 at 2:46
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    $\begingroup$ Bourbaki's Lie groups and Lie algebras, chapter VI, §4, exercise 1. $\endgroup$
    – abx
    Feb 4, 2016 at 5:25
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    $\begingroup$ another description of the simple subgroup is as follows: there are two classes of maximal (dimension 4) totally isotropic, w.r.t. to the form preserved by the group, subspaces (each class of size 135). The simple subgroup preserves these classes, elements outside mix them. $\endgroup$ Feb 11, 2016 at 17:53

2 Answers 2

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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.

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    $\begingroup$ Incidentally, the notation issues with finite simple groups of Lie type in type $D$ are bemoaned at en.wikipedia.org/wiki/Group_of_Lie_type#Notation_issues. $\endgroup$
    – Marty
    Feb 4, 2016 at 5:57
  • $\begingroup$ Thanks. The group called $\mathrm{O}_8^+(2)$ in the original question is not $\mathrm{SO}(\bar{\Omega}),N)$ but $\mathrm{O}(\bar{\Omega},N)$. My question was thus in part whether there's a 2-1 homomorphism from $W$ to $\mathrm{O}(\bar{\Omega},N)$. You're saying there's a homomorphism, but not whether it's 2-1. But I'm pretty sure it's 2-1 now. $\endgroup$
    – John Baez
    Feb 5, 2016 at 7:01
  • $\begingroup$ Yep - the kernel of $W \rightarrow O(\bar \Omega, N)$ is $\{ \pm 1 \}$. $\endgroup$
    – Marty
    Feb 5, 2016 at 9:58
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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.

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