Exercise 13, § 5

Let $(\mathfrak{g},\mathfrak{h},B,(X_{\alpha})_{\alpha\in B})$ be a framed semi-simple Lie algebra, R the root system of $(\mathfrak{g},\mathfrak{h}),\Delta$ the corresponding Dynkin graph, and $\Phi$ a subgroup of

$^{13}$For more details on this exercise, see: J. TITS, Normalisateurs de tores. I. Groupes

de Coxeter étendus, J. Alg., Vol. IV (1966), pp. 96-116.

Aut(R$,B$) $=$ Aut($\Delta$ ).

If $s\in \Phi$, extend $s$ to an automorphism of $\mathfrak{g}$ by the conditions

$s(X_{\alpha}) =X_{s\alpha}$, and $s(H_{\alpha}) =H_{s\alpha}$ for all $\alpha \in B$, cf. Prop$.1$.

We thus identify $\Phi$ with a subgroup of Aut($\mathfrak{g},\mathfrak{h}$); denote by $\widetilde{\mathfrak{g}}$ (resp. $\widetilde{\mathfrak{h}}$) the subalgebra of $\mathfrak{g}$ (resp. $\mathfrak{h}$) consisting of the elements invariant under $\Phi$.

a) Let $\alpha \in B$, and let $X =\Phi .\alpha$. Show, by using the Plates in Chap. VI, that only two cases are possible:

(i) every element of X distinct from $\alpha$ is orthogonal to $\alpha$;

(ii) there exists a unique element $\alpha '$ of X-- $\{\alpha \}$ that is not orthogonal to $\alpha$, and $n(\alpha , \alpha ') =n(\alpha ', \alpha ) =-1$.

b) Let $i$ be the restriction map $\mathfrak{h}^*\rightarrow \widetilde{\mathfrak{h}}^*$, and let $\widetilde{B} =i(B)$; the map $B\rightarrow \widetilde{B}$ identifies $\widetilde{B}$ with $B/\Phi$. Show that $\widetilde{\mathfrak{g}}$ is a semi-simple Lie algebra, that $\widetilde{\mathfrak{h}}$ is a splitting Cartan subalgebra of it, and that $\widetilde{B}$ is a basis of $R(\widetilde{\mathfrak{g}},\widetilde{\mathfrak{h}})$. (Observe that $\widetilde{B}$ is contained in $R(\widetilde{\mathfrak{g}},\widetilde{\mathfrak{h}})$, and that every element of $R(\widetilde{\mathfrak{g}},\widetilde{\mathfrak{h}})$ is a linear combination of elements of $\widetilde{B}$ with integer coefficients of the same sign.) If $\widetilde{\alpha}\in \widetilde{B}$, the corresponding inverse root $H_{\widetilde{\alpha}}\in \widetilde{\mathfrak{h}}$ is given by

$H_{\widetilde{\alpha}}=\sum_{i(\alpha)=\widetilde{\alpha}}H_{\alpha}$ in case (i) of a)

$H_{\widetilde{\alpha}}= 2\sum_{i(\alpha)=\widetilde{\alpha}}H_{\alpha}$ in case (ii) of a),

where the summation is over those elements $\alpha \in B$ such that $i(\alpha ) = \widetilde{\alpha}$. If $\beta \in B$ has image $\widetilde{\beta}\in \widetilde{B}$, then

$n( \widetilde{\beta ,}\widetilde{\alpha}) =\sum_{i(\alpha)=\widetilde{\alpha}}n(\beta , \alpha )$ in case (i)

$n( \widetilde{\beta ,}\widetilde{\alpha}) = 2\sum_{i(\alpha)=\widetilde{\alpha}}n(\beta , \alpha )$ in case (ii).

Deduce how the Dynkin graph of $R(\widetilde{\mathfrak{g}},\widetilde{\mathfrak{h}})$ is determined starting from the pair $(\Delta , \Phi )$.

c) Show that, if $\mathfrak{g}$ is simple, so is $\widetilde{\mathfrak{g}}$.

If $\mathfrak{g}$ is of type $A_l,l\geq 2$, and $\Phi$ is of order 2, then $\widetilde{\mathfrak{g}}$ is of type $B_{l/2}$ if $l$ is even, and of type $C_{(l+1)/2}$ if $l$ is odd.

If $\mathfrak{g}$ is of type $D_l,l\geq 4$, and $\Phi$ is of order 2, then $\widetilde{\mathfrak{g}}$ is of type $B_{l-1}$.

If $\mathfrak{g}$ is of type $D_4$, and $\Phi$ is of order 3 or 6, then $\widetilde{\mathfrak{g}}$ is of type $G_2$.

If $\mathfrak{g}$ is of type $E_6$, and $\Phi$ is of order 2, then $\widetilde{\mathfrak{g}}$ is of type $F_4$.