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Let $\mathfrak{g}$ be a complex semisimple Lie algebra, $\Delta$ its root system contained in $\mathfrak{t}^{\vee}$ for a Cartan sub-algebra $\mathfrak{t}$ of $\mathfrak{g}$. Let $W$ be its Weyl group. Then we (might) have the following short exact sequence of groups: $$ 1\rightarrow W \rightarrow Aut(\Delta) \rightarrow E \rightarrow 1$$ where $Aut(\Delta)\subset GL(\mathfrak{t}^{\vee})$ is the group of symmetries of $\Delta$, and $E$ (should) be the group of symmetries of the associated Dynkin diagram.

It is possible to see how the map $Aut(\Delta) \rightarrow E$ is defined from the perspective of simple roots (let them be $\{\alpha_1,\ldots,\alpha_k\}$): let $\sigma\in Aut(\Delta)$, then $\{\sigma(\alpha_1),\ldots,\sigma(\alpha_k)\}$ is a set of simple roots with respect to a choice of positive roots which corresponds to its dominant Weyl chamber. By composing $\sigma$ with the unique Weyl element mapping this chamber to the original dominant Weyl chamber, we get a permutation of $\{\alpha_1,\ldots,\alpha_k\}$. One can see that this permutation preserves the Cartan matrix and thus defines an element of $E$.

What I would like to know is whether there is an alternative way to define the above map from the perspective of dominant fundamental weights, namely, let $\{\lambda_1,\ldots,\lambda_k\}$ be the corresponding fundamental weights, then I observe that any element $\sigma\in Aut(\Delta)$ induces a permutation of the Weyl orbits generated by the $\lambda_i$'s. So we almost get the desired map except we haven't checked this permutation preserves the Cartan matrix. This raises the following question:

Can these Weyl orbits themselves (as polyhedra in the Euclidean space) tell us the Cartan matrix?

As an example, $A_3$, these orbits consist of two congruent tetrahedra and one octahedron. We know that there is no edge joining the two tetrahedra in the Dynkin diagram. Can we see this fact from their configuration?

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  • $\begingroup$ Have you been able to treat Lie types $B_\ell, C_\ell$? $\endgroup$ – Jim Humphreys Jan 15 at 1:13
  • $\begingroup$ On the surface there seems to be no clear reason to work over $\mathbb{C}$, or even in characteristic 0. $\endgroup$ – Jim Humphreys Jan 15 at 1:16
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    $\begingroup$ Intuitively it seems you should be able to do the following: we have a bunch of vertices for all these polyhedra; let's try to make a complete fan out of these where these vertices correspond to the rays of the fan; the rule is just, form a cone for a given set of vertices whenever there is no other vertex in the cone they generate; if this works it should give you the Coxeter arrangement. So then pick any chamber in the Coxeter arrangement and the vertices generating that chamber are your fundamental weights. Inner products of fundamental weights give you the (inverse) of the Cartan matrix $\endgroup$ – Sam Hopkins Jan 15 at 2:30
  • $\begingroup$ Another way to say what I'm trying to say is that the fundamental weights should be the vertices forming the "sharpest" angle: they span the whole space and have no other vertices in the cone they generate. $\endgroup$ – Sam Hopkins Jan 15 at 2:40

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