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I recently needed to know which circles $S$ in a maximal torus $T^6$ of the compact exceptional group $E_6$ yield one-dimensional subspaces $\mathfrak s$ of the Lie algebra $\mathfrak t^6$ that are reflected by some element of the Weyl group $W_{E_6}$.

I found that such ($-1$)-eigenspaces were precisely those contained in some four-dimensional subspace $\mathfrak t^4$ of $\mathfrak t^6$ spanned by four mutually orthogonal coroots. Moreover, there are exactly forty-five such $\mathfrak t^4$, each tangent to a maximal torus of a $\mbox{Spin}(8)$ subgroup of $E_6$.

From there, there's an incidental corollary that one of these injections $\mathfrak t^4 \to \mathfrak t^6$ induces an injection of Weyl groups $W_{F_4} \to W_{E_6}$, which just follows from the orbit–stabilizer theorem.

It seems very likely these facts were known. Does anyone know a reference?

Versions of this question have previously been asked here:

https://math.stackexchange.com/questions/884275/the-weyl-group-of-e-6-acting-on-embedded-circles https://math.stackexchange.com/questions/1144878/an-injection-of-weyl-groups

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    $\begingroup$ I would suggest this is very much related to the fact that the longest element of the Weyl group of type E6 is $-\sigma$ where $\sigma$ is the automorphism of the root system induced by the unique non-trivial automorphism of the Dynkin diagram. Thus the longest element acts as $-1$ on anything fixed by the diagram automorphism. The fixed point subgroup under $\sigma$ is the group obtained by the diagram fold, which is just the Weyl group of type F4. $\endgroup$
    – Jay Taylor
    Apr 19, 2015 at 21:46
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    $\begingroup$ This suggests to me (I don't really understand diagram folding well, so I apologize if there's something obviously wrong in what follows) that the injection of Weyl groups should follow from folding. It also seems from your description that folding implies it should be a sufficient condition on $\mathfrak s$ that it lie within the tangent space to a maximal torus of an $F_4$ subgroup. I don't understand why necessity should follow from this line of reasoning, though, or indeed if you are claiming it should. $\endgroup$
    – jdc
    Apr 19, 2015 at 23:59

2 Answers 2

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EDIT II: Sorry to bump this again but the answer to this question can be phrased entirely in terms of finite Coxeter groups and doesn’t depend at all on the fact that we’re dealing with $\textsf{E}_6$, so it seemed best to write the answer that way.

Let $V$ be a finite dimensional real euclidean vector space and let $\Phi \subseteq V$ be a root system in the sense of §1.2 in Humphrey’s “Reflection groups and Coxeter groups”. For convenience we will assume that $\Phi$ spans $V$. We will denote by $W = \langle s_{\alpha} \mid \alpha \in \Phi \rangle$ the corresponding reflection group and we choose a set of simple roots $\Delta \subseteq \Phi$.

If $w_0 \in W$ is the longest element, with respect to $\Delta$, then $-w_0(\Delta) = \Delta$ and hence $-w_0$ defines an automorphism $\sigma : V \to V$ such that $\sigma(\Delta) = \Delta$.

We will denote by $\mathcal{S}$ the set of all 1-dimensional subspaces $A \subseteq V$ such that $w(v) = -v$ for some $w \in W$ and any $v \in A$. Clearly $W$ acts on the set $\mathcal{S}$ because we have

\begin{equation*} w(v) = -v \Leftrightarrow xwx^{-1}(x(v)) = -x(v) \end{equation*}

for any $x \in W$. Let $D = \{\lambda \in V \mid (\lambda,\alpha) \geqslant 0$ for all $\alpha \in \Delta\}$ be the closure of the fundamental Weyl chamber of $W$, where $(-,-) : V \times V \to \mathbb{R}$ is the euclidean form on $V$. As any element in $V$ is conjugate under $W$ to a unique element in $D$, i.e., $D$ is a fundamental domain for the action of $W$, it suffices to determine for which $v \in D$ we have $\mathbb{R}v \in \mathcal{S}$.

Assume $v \in D$ is such that $w(v) = -v$ for some $w \in W$ then we have

\begin{equation*} w_0w(v) = -w_0(v) = \sigma(v). \end{equation*}

Now assume $\lambda \in D$ then $(\sigma(\lambda),\sigma(\alpha)) = (\lambda,\alpha) \geqslant 0$ for all $\alpha \in \Delta$. As $\sigma(\Delta) = \Delta$ this shows that $\sigma(D) = D$. Hence $w_0w(v) \in D$ but as $v \in D$ this implies $w_0w(v) = v$ and $w_0w = 1$. In particular, we have $\sigma(v) = v$ and $w = w_0$.

Consider the subgroup $H = \{w \in W \mid w_0ww_0 = w\}$ fixed under conjugation by $w_0$. If $W/H$ are the cosets of $H$ in $W$ then it is clear from the above discussion that we have

\begin{equation*} \mathcal{S} = \bigsqcup_{x \in W/H} x(\mathcal{S}^{\sigma}) \end{equation*}

where $\mathcal{S}^{\sigma} = \{A \in \mathcal{S} \mid \sigma(A) = A\}$.

Firstly, if $w_0 = -1$ then it is clear that $\mathcal{S}$ simply contains all 1-dimensional subspaces of $V$. Now, let us consider your special situation of $\textsf{E}_6$. We number the simple roots $\Delta = \{\alpha_1,\dots,\alpha_6\}$ as in Bourbaki. In this case $\sigma(\alpha_i) = \alpha_{\rho(i)}$ where $\rho \in \mathfrak{S}_6$ is the permutation $(1,6)(3,5)$. It is well known that the subgroup $H$ is isomorphic to the Weyl group $W_{\textsf{F}_4}$, see for instance 13.3.3 of Carter's "Simple groups of Lie type". Comparing the orders of the Weyl groups of type $\textsf{E}_6$ and $\textsf{F}_4$ we see that $|W/H| = 45$, which explains your 45 tori.

I am not aware of a reference for this fact but in the end it's reasonably straight forward.

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  • $\begingroup$ I'm not sure I understand your answer. Regarding the first paragraph, I don't need my element to be in the $(-1)$-eigenspace of a reflection per se; if there's some other element of $W$ that reflects it, I'm equally happy with it. Hence I'm not sure asking for coroots suffices. Am I misunderstanding something? $\endgroup$
    – jdc
    Apr 20, 2015 at 19:12
  • $\begingroup$ (con'td) That said, I did understand that the coroots fixed by $-\sigma$ are the $F_4$ system claimed, and that lying in or being conjugate the span suffices. (This is roughly how I found the 45 in the first place, except that I viewed it as being spanned by a $D_4$ sublattice.) What I don't understand is the theoretical reason that being conjugate to something in the $(-1)$-eigenspace of $w_0$ is necessary. (I personally wound up manually checking $(-1)$-eigenspaces of elements of orders 4 and 8 with GAP to conclude necessity, but this was psychologically unsatisfactory.) $\endgroup$
    – jdc
    Apr 20, 2015 at 19:16
  • $\begingroup$ I see your problem now. I hadn't realised exactly what you were asking for. I still think it shouldn't be too difficult. It should follow from the fact that $-1 = w_0\sigma$ but I'll have to think about it. Sorry it's not yet a complete answer. $\endgroup$
    – Jay Taylor
    Apr 23, 2015 at 14:11
  • $\begingroup$ No worries! I think it's most of the way to an answer, and in any event, it's just nice to get feedback from someone. I tend to ask questions MSE doesn't care about but worry they're too low-level for MO. $\endgroup$
    – jdc
    Apr 23, 2015 at 23:37
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    $\begingroup$ I edited my answer, hopefully this is now complete. Your idea of using the closure of the fundamental Weyl chamber was the right call. $\endgroup$
    – Jay Taylor
    Apr 24, 2015 at 13:56
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I'll just add a few comments to what Jay has already said, in community-wiki style.

1) Most of these ideas have been developed over the past century, from E. Cartan onward, but at first in the Lie group context and later in more algebraic settings. It's always difficult to say what is "new", but as far as I know, no single reference for your formulation can just be quoted verbatim. Even so, there are quite a few relevant treatments in the literature.

2) The Weyl groups themselves have been studied in many guises, as indicated in section 2.12 of my book on reflection groups along with the sources (such as the Atlas of Finite Groups) mentioned there. For example, $W(E_6)$ can be realized as the automorphism group of the famous 27 lines on a cubic surface. Bourbaki's tables in Chapters 4-6 of Groupes et algebres de Lie provide considerable detail about the roots and Weyl groups of each irreducible type.

3) In particular, $|W(E_6)| = 2^7 \, 3^4 \, 5$, whereas $|W(F_4)| = 2^7 \, 3^2$. The fact that the latter group is embedded in the former one (with index 45) via "folding" relative to an automorphism of the Dynkin diagram is developed, as Jay notes, in Carter's 1972 Simple Groups of Lie Type, especially 13.1-13.3. (But be careful about his numbering of vertices in Dynkin diagrams.) There are similar treatments elsewhere, for example in Steinberg's 1967-68 Yale lecture notes on Chevalley groups: see $\S11$ and especially page 176 for $E_6$. These notes are still available online here. Of course, the emphasis in such references is on the finite Chevalley groups, but their Weyl groups play an essential role throughout.

4) Counting configurations relative to finite group actions is very often done most efficiently by looking at orbits and isotropy groups, so for example the number 45 is not accidental here. But it should be viewed as a consequence of the Weyl group embedding, not the other way around.

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