Remark 2

Let X be the set of pairs $(\mathfrak{h},B)$ where $\mathfrak{h}$ is a splitting Cartan subalgebra of $\mathfrak{g}$ and B a basis of the root system of $(\mathfrak{g},\mathfrak{h})$. If $x= (\mathfrak{h},B)$ and $x'= (\mathfrak{h}',B')$ are two elements of X, there exists $s\in$ Aut$_0(\mathfrak{g})$ that transforms $x$ into $x'$ (Prop. 5), and the restriction $s_{x',x}$ of $s$ to $\mathfrak{h}$ does not depend on the choice of $s$ (Remark 1). In particular, $s_{x'',x'}\circ s_{x',x}=s_{x'',x}$ if $x, x', x''\in X$, and $s_{x,x}= 1$. The set of families $(h_x)_{x\in X}$ satisfying the conditions

$a)h_x\in \mathfrak{h}$ if $x= (\mathfrak{h},B)$

$b)s_{x',x}(h_x) =h_{x'}$ if $x, x'\in X$

is in a natural way a vector space $\mathfrak{h}(\mathfrak{g})$ which we sometimes call the canonical Cartan subalgebra of $\mathfrak{g}$. For $x= (\mathfrak{h},B)$ and $x'= (\mathfrak{h}',B'),s_{x',x}$ takes B to $B'$, and hence the root system of $(\mathfrak{g},\mathfrak{h})$ to that of $(\mathfrak{g},\mathfrak{h}')$; it follows that the dual $\mathfrak{h}(\mathfrak{g})^*$ of $\mathfrak{h}(\mathfrak{g})$ is naturally equipped with a root system $R(\mathfrak{g})$ and with a basis $B(\mathfrak{g})$ of $R(\mathfrak{g})$. We sometimes say that $R(\mathfrak{g})$ is the canonical root system of $\mathfrak{g}$ and that $B(\mathfrak{g})$ is its canonical basis. The group Aut($\mathfrak{g}$) operates on $\mathfrak{h}(\mathfrak{g})$ leaving $R(\mathfrak{g})$ and $B(\mathfrak{g})$ stable; the elements of Aut($\mathfrak{g}$) that operate trivially on $\mathfrak{h}(\mathfrak{g})$ are those of Aut$_0(\mathfrak{g})$.