9
$\begingroup$

Is there any finite 8-dimensional point group which contains the $E_8$ Coxeter group as a subgroup other than $E_8$ itself?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
19
$\begingroup$

No.

Let $G$ be a finite subgroup of $GL_n(\mathbb R)$ containing $W(E_8)$ as a subgroup. Because $G$ is compact, $G$ must preserve a symmetric positive definite form on $\mathbb R^8$. Since $W(E_8)$ preserves a unique such form, it must be that one.

Let $H$ be the largest subgroup of $G$ generated by reflections. Then $H$ contains $W(E_8)$ and thus is an irreducible reflection group, hence a Coxeter group. Examining the table of Coxeter groups and looking for entries in dimension $8$, there are four possibilities: $W(A_8), W(B_8), W(D_8), W(E_8)$. Because $W(E_8)$ has the highest order of these, we must have $H = W(E_8)$.

Now, by construction, $H$ is a normal subgroup of $G$, so $G$ normalizes $W(E_8)$, and hence $G$ is contained in the automorphism group of the $E_8$ root system. Because the Coxeter-Dynkin diagram $E_8$ has no nontrivial automorphisms, this is $W(E_8)$ and so $G= W(E_8)$, as desired.

Improve this answer
$\endgroup$
10
  • $\begingroup$ In the last step, why should $G$ be a semidirect product? Even if it is, why should it inject into $\mathrm{Aut}E_8$? Also, $E_8$ does have a non-trivial outer automorphism (taking reflections to their negation). Diagram automorphisms induce group automorphisms, but are not always outer nor do they always exhaust the outer automorphisms. $\endgroup$
    – Grant B.
    Commented Feb 11, 2021 at 20:38
  • 1
    $\begingroup$ @GrantB. I said nothing about semidirect products. We can describe the $E_8$ root system as the set consisting of, for each reflection in $E_8$, the two vectors of length $2$ which are eigenvectors of the reflection with eigenvalue $1$. This makes it clear that any $G$ which normalizes $E_8$ stabilizes the $E_8$ root system, and hence, because there are no Dynkin diagram automorphisms, lies in $W(E_8)$. $\endgroup$
    – Will Sawin
    Commented Feb 11, 2021 at 21:58
  • 1
    $\begingroup$ @GrantB. Said to work in the generality of Coxeter groups, rather than just ADE type root systems: Consider the division of $\mathbb R^n$ by hyperplanes fixed by the reflections in a Coxeter group $H$. This splits $\mathbb R^n$ into a union of alcoves, on which $H$ and $N(H)$ (the normalizer of $H$ in $O(N)$, not its outer automorphism group) act. But $H$ acts transitively on alcoves because we can use the appropriate reflection to connect adjacent chambers, so every element of $N(H)$ is an element of $H$ times a stabilizer of your favorite alcove - i.e. an automorphism of the Coxeter diagram. $\endgroup$
    – Will Sawin
    Commented Feb 11, 2021 at 22:05
  • 1
    $\begingroup$ @GrantB. Negation IS an element of the Coxeter group $E_8$. Specifically, it is the longest element. In general, negation is contained in a finite Coxeter group if and only if all of its exponents (see encyclopediaofmath.org/wiki/Coxeter_group ) are odd. $\endgroup$
    – David E Speyer
    Commented Feb 12, 2021 at 1:28
  • 1
    $\begingroup$ @David Right, I agree that $-1=w_0$ is in $E_8$, but the automorphism given by $s_i\mapsto -s_i$ is still outer, since for instance the trace of each reflection changes sign. $\endgroup$
    – Grant B.
    Commented Feb 12, 2021 at 5:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .