Timeline for Can $E_8$ be enlarged?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2021 at 9:30 | history | edited | F. C. | CC BY-SA 4.0 |
fix typo in Coxeter name
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| Feb 12, 2021 at 13:36 | comment | added | David E Speyer | @GrantB. Okay, now I understand. You are right (and Will is also right). | |
| Feb 12, 2021 at 5:58 | comment | added | Grant B. | @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. | |
| Feb 12, 2021 at 5:46 | comment | added | Grant B. | @Will Ahh, I missed the phrase "root system" when I was reading, my mistake! Your follow-up clarifies the role of the normalizer in O(n), thank you! | |
| Feb 12, 2021 at 1:41 | comment | added | David E Speyer | One way to see this is the ring of invariants is generated in degrees $e_1+1$, $e_2+1$, ..., $e_n+1$, where the $e_j$ are the exponents, so "all $e_j$ odd" is equivalent to "all invariant functions are in even degree" is equivalent to "all invariant functions are invariant for $-1$" is equivalent to "$-1$ is in the group". | |
| Feb 12, 2021 at 1:39 | comment | added | David E Speyer | I am sure we basically agree, but I don't quite follow how you are saying it. If $-1$ is NOT in the Coxeter group, then $(-1) \circ w_0$ permutes the simple roots nontrivially and gives rise to a diagram automorphism. Thus, if there are no diagram automorphisms (as in the case of $E_8$), we can be sure that $-1$ is in the group. But there are cases which have diagram automorphisms, and yet $-1$ is in the group: This happens for $D_{2k}$. I thought I'd give the exact criterion for $-1$ to be in the group. | |
| Feb 12, 2021 at 1:34 | comment | added | Will Sawin | @DavidESpeyer The issue is that every Coxeter group containing negation has an outer automorphism which sends every reflection to minus that reflection, and this isn't an automorphism coming from the Coxeter group or diagram automorphisms (except possibly in dimension 2), because it doesn't even preserve the characteristic polynomials of the reflections. Of course, this is not so relevant here, because we're interested in automorphisms coming form linear transformations only. | |
| Feb 12, 2021 at 1:28 | comment | added | David E Speyer | @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. | |
| Feb 11, 2021 at 22:05 | comment | added | Will Sawin | @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. | |
| Feb 11, 2021 at 21:58 | comment | added | Will Sawin | @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 $W(E_8)$. | |
| Feb 11, 2021 at 20:38 | comment | added | Grant B. | 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. | |
| Feb 11, 2021 at 13:50 | history | answered | Will Sawin | CC BY-SA 4.0 |