Are there any proper subgroups of the Coxeter group $W(E_8)$ which are also proper overgroups of $W(A_8)$, other than $\text{Aut}(A_8)$?
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2$\begingroup$ @YCor The lattice of vectors in $ ( (1/3) \mathbb Z)^9$ which sum to $0$ and are congruent mod $1$ is a model of $E_8$, and its symmetry group contains $W(E_8) = S_9$ permuting the entries. $\endgroup$– Will SawinApr 1, 2021 at 18:02
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1$\begingroup$ Equivalent to @WillSawin's answer: $E_8$ Lie group contains an element of order $3$ whose centralizer is $SU(9)/\mathbb{Z}_3$. $\endgroup$– Theo Johnson-FreydApr 1, 2021 at 18:58
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3$\begingroup$ @YCor. Here's how you embed the $A_8$ root system into the $E_8$ root system. Consider the $8$ simple roots of $E_8$ together with the negative of the highest root. Together, they form a copy of the affine $E_8$ Dynkin diagram. Now, there's an obvious copy of the $A_8$ Dynkin diagram inside the affine $E_8$ Dynkin diagram. The same method produces lots of same-rank inclusions between Lie algbras. For example $A_4\times A_4$ sits inside $E_8$ . $\endgroup$– André HenriquesApr 1, 2021 at 19:57
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2$\begingroup$ @AndréHenriques's answer is an example of Borel–de Siebenthal theory: maximal-rank subgroups are classified by sub-root systems of the extended root system $\widetilde{E_8}$. Deleting $\alpha_2$ (in the Bourbaki notation) from that extended system gives $A_8$. $\endgroup$– LSpiceApr 2, 2021 at 2:36
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1$\begingroup$ @LSpice Sometimes you need to do the trick of passing the the extended Dynkin diagram more than once: for example $A_2\times A_2\times A_2\times A_2$ sits inside $E_8$. $\endgroup$– André HenriquesApr 3, 2021 at 23:07
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1 Answer
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One approach is to calculate the orbits of $W(A_8)$ on $W(E_8) / W(A_8)$.
I claim these orbits have sizes $1, 1, 84, 84, 560, 560, 630$.
Given this claim, it's straightforward to check. For any intermediate subgroup $G$, $G / W(A_8)$ must have an order a divisor of $1920$ and must be a union of these orbits, including $1$. We can't write $1920/2= 960$ as a sum of these numbers because $630+1+1+84+84$ is too small but including two of $560$, $560$, $630$ would be too big. We can't write $1920/3=640$ for similar reasons. Because $1920/4 =480<560$, the only remaining possibilities are $1$, $2$, $85$, $86$, $169$, $170$ and none of those is a divisor of $1920$ except $1$ and $2$, which correspond to $W(A_8)$ and its normalizer.
The way I calculated this involves observing that because the inclusion $W(A_8) \subset W(E_8)$ comes from viewing the lattice $A_8$ as an index $3$ sublattice of $E_8$, the kernel of a map $E_8 \to \mathbb Z/3$ arising by dot product with an element of $E_8$ mod $3$, we can represent $W(E_8)/W(A_8)$ as a $W(E_8)$-orbit inside $E_8$ mod $3$ — specifically, the elements with norm congruent to $2$ mod $3$ that aren't roots.
I then found all these elements in a model of $E_8$ on which $W(A_8)=S_9$ acts, i.e. vectors in $((1/3) \mathbb Z)^9$ that sum to zero, and calculated the $S_9$ orbits. They are
- $630$ permutations of $(0,1,1,1,1,-1,-1,-1,-1)$
- $1$ permutation of $(1/3,1/3,1/3,1/3,1/3,1/3,1/3,1/3,1/3)$ (after subtracting $3$ from one entry — mod $3E_8$, it doesn't matter which one)
- $84$ permutations of $(1/3,1/3,1/3, -2/3,-2/3,-2/3,-2/3, -2/3, -2/3)$ (after adding $3$ to one entry)
- $560$ permutations of $(4/3, 4/3, 4/3,-2/3,-2/3,-2/3,1/3,1/3, 1/3) $ (after subtracting $3$ from one entry)
- and the negations of the last three orbits.
answered Apr 1, 2021 at 18:46