# Can $E_8$ be enlarged?

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

• What do you mean by symmetry group? Surely you don't want to include $O(8)$, the symmetry group of the sphere. Do you mean a finite subgroup of the orthogonal group? – Will Sawin Feb 11 at 3:29
• Oops. I meant to add that but I must’ve forgotten. I’ve made an edit and fixed it. – Daniel Sebald Feb 11 at 3:51
• Possibly related: mathoverflow.net/questions/37136/… – Sam Hopkins Feb 11 at 3:55
• What is a finite symmetry group? A finite reflection group? A finite subgroup of O(8)? – Moishe Kohan Feb 11 at 4:46
• Please consider adding the tag "coxeter-groups" to this question. – Nathan Reading Feb 11 at 15:42

## 1 Answer

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.

• 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. – Grant B. Feb 11 at 20:38
• @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 $W(E_8)$. – Will Sawin Feb 11 at 21:58
• @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. – Will Sawin Feb 11 at 22:05
• @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. – David E Speyer Feb 12 at 1:28
• @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. – Grant B. Feb 12 at 5:58