$L$ 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 $H$. ]

  • $\begingroup$ 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

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 $\sigma(x)\in L_{\beta}$.


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 $\alpha\circ \sigma^{-1} = \sigma(\alpha)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.