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$L$ a simple Lie algebra over $\mathbb{C}$ (finite dim.), $H$ a maximal toral, $\Phi$ the root system relative to $H$.

Then $L$ has Cartan decomposition $$L=H\oplus \amalg_{\alpha\in\Phi} L_{\alpha}.$$

In the discussion (and for question) fix $\sigma$ an automorphism in Weyl group of $\Phi$.

Then $\sigma$ induces an automorphism of $H$ [See justification below].

Thus think of $\sigma$ as automorphism of $H$ obtained in a natural way.

There is a way to construct an extension of $\sigma$ to an automorphism of $L$. It is described in Humphreys' Lie algebra, Section 14.

Q. If $\sigma$ takes root $\alpha$ to $\beta$ (and hence $h_{\alpha}$ to $h_{\beta}$ in $H$), then in an extension of $\sigma$ to automorphism of $L$, is it necessary that $\sigma$ should take $L_{\alpha}$ to $L_{\beta}$?


[Passing from automorphism of $\Phi$ to automorphism of $H$:

given any root $\alpha\in\Phi$; since Killing form is non-generate on $L$ as well as $H$, so the map $H^*\rightarrow H$ induced by Killing form is isomorphism. Hence for $\alpha\in \Phi$ - which is in fact an element of $H^*$ - there is unique $t_{\alpha}\in H$ such that $\alpha$ looks like $\kappa(t_{\alpha}, -)$. For this $t_{\alpha}$, set $h_{\alpha}:=\frac{2t_{\alpha}}{\kappa(t_{\alpha}, t_{\alpha})}$ So each $\alpha$ determines a unique $h_{\alpha}\in H$. If $\sigma(\alpha)=\beta$, define $\sigma(h_{\alpha})=h_{\beta}$. Since $H$ is abelian Lie algebra, this is clearly automorphism of $H$. ]

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  • $\begingroup$ In the Cartan decomposition, I would always write $L=H\oplus \bigoplus_{\alpha\in\Phi} L_{\alpha}$. It really is a sum, not a disjoint union or something: Sums of root space vectors are vectors too (although annoyingly, they are not in root spaces anymore in general.). $\endgroup$ – Torsten Schoeneberg Apr 27 '18 at 22:29
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Yes! Fix any $h_{\gamma}$ corresponding to a root $\gamma\neq \beta$ and assume that $\sigma:h_{\alpha}\mapsto h_{\beta}$ and $\sigma:h_{\delta}\mapsto h_{\gamma}$. Let $x\in L_{\alpha}$. The Killing form is invariant under automorphisms of the Lie algebra. We easily verify that $$ \beta(h_{\gamma})=\frac{\beta(t_{\beta})}{\alpha(t_{\alpha})}\alpha(h_{\delta}). $$ Since $\sigma$ is induced from an isometry of the root system (being an element of the Weyl group), it follows that $$\beta(t_{\beta})=\kappa(t_{\beta},t_{\beta})=\langle \beta,\beta \rangle=\langle \alpha,\alpha \rangle=\alpha(t_{\alpha}).$$ so that $$\beta(h_{\gamma})\sigma(x)=\frac{\alpha(t_{\alpha})}{\beta(t_{\beta})}\beta(h_{\gamma})\sigma(x)=\alpha(h_{\delta})\sigma(x)=\sigma([h_{\delta},x])=[h_{\gamma},\sigma(x)]\;.$$ For $\gamma=\beta$, we have $$\beta(h_{\beta})\sigma(x)=2\sigma(x)=\alpha(h_{\alpha})\sigma(x)=[h_{\beta},\sigma(x)].$$ Extending by linearity shows that indeed $\sigma(x)\in L_{\beta}$.

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The other answer is good, I just want to give a slightly modified version which avoids using the Killing form.

$\sigma$ being an automorphism of $L$ means by definition that (among other things) $\sigma$ is linear and

$$[\sigma(x), \sigma(y)] = \sigma([x,y])$$

for all $x,y \in L$. Now let $0\neq e_\alpha \in L_\alpha$. Then for all $h \in H$

$$ [h, \sigma(e_\alpha)] = [\sigma(\sigma^{-1}(h)), \sigma(e_\alpha)] = \sigma([\sigma^{-1}(h),e_\alpha])= \sigma(\alpha(\sigma^{-1}(h)) \cdot e_\alpha) = \alpha(\sigma^{-1}(h)) \cdot \sigma(e_\alpha)$$

meaning that $H$ acts on $\sigma(e_\alpha)$ via the weight $\alpha\circ \sigma^{-1}$. Now check that by how you defined $\sigma$ on $H$, you have $\alpha\circ \sigma^{-1} = \sigma(\alpha)$.

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