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My question is related to this post. I am trying to compare the Weyl groups associated to root systems of type $D_2$ and $A_3$ respectively.

I know that the simple roots of $D_2$ are $e_1-e_2$ and $e_1+e_2$, and that the simple roots of $A_3$ are $e_1-e_2$, $e_2-e_3$ and $e_3-e_4$. The Weyl group $W'$ of the former root system can be written as $\{1,-1,s,-s\}$ where $-1$ is the sign change and $s$ is the permutation. The Weyl group of the latter is just $W\cong S_4$.

Is there a canonical way of embedding $W'$ in $W$? My question comes from the fact that I want to study $\text{SO}_4\subset \text{GL}_4$, where $\text{SO}_4$ is the orthogonal group where each matrix has determinant $1$. As there is an inclusion of reductive groups, there should be an inclusion of Weyl groups.

Many thanks in advance.

EJB
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  • " As there is an inclusion of reductive groups, there should be an inclusion of Weyl groups." -- I think that is a far more subtle issue than one might think. – Torsten Schoeneberg Jul 28 '23 at 03:47
  • I can image that I should be more careful. But what can go wrong in this specific example? The Weyl group is isomorphic to $N_G(T)/T$ where $T$ is a maximal torus of $G$. In this specific example we have an inclusion of normalisers $N_{G'}(T') \subset N_G(T)$ and tori $T'\subset T$ where $G'=\text{SO}_4$ and $G=\text{GL}_4$. Does the natural map between these quotients not give us the inclusion? – EJB Jul 28 '23 at 12:18
  • One idea that I had recently is that $W\cong S_4$ acts on the set ${e_1,e_2,e_3,e_4}$, whereas $W'$ acts on the set ${\pm e_1,\pm e_2}$. If we make the naive and silly relabelling $e_3:=-e_1$ and $e_4:=-e_2$ we can naturally see $W'$ as a subset of $W$. Namely, now $W'\cong \langle (12)(34),(14)(23)\rangle$. Does this make any sense? Does this depend at all on the choice of relabelling? – EJB Jul 28 '23 at 12:45
  • Why is $N_{G'}(T') \subset N_G(T)$? (I.e. why would something that normalizes the smaller torus automatically normalize the big one?) That would certainly not be true in general. If it is so here, then why don't you write all those groups down explicitly with matrices, and see what map gets induced? – Torsten Schoeneberg Jul 28 '23 at 21:09
  • In this situation we have the inclusion because of the definition of $T'$. Many thanks for your point, because now I see that the classes of the matrices $\begin{pmatrix} 0&1&0&0\ 1&0&0&0\ 0&0&0&1\ 0&0&1&0 \end{pmatrix}$ and $\begin{pmatrix} 0&0&1&0\ 0&0&0&1\ 1&0&0&0\ 0&1&0&0 \end{pmatrix}$ generate $W'$. However, how do I know that these are the elements of length $1$, and their product is of length $2$? – EJB Aug 02 '23 at 13:43

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