$L$ 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 $L=H\oplus \amalg_{\alpha\in\Phi} L_{\alpha}.$

For a root $\alpha$ let $\sigma_{\alpha}$ denote the reflection $\beta\mapsto \beta- \langle \beta,\alpha\rangle\alpha$; it is an automorphism of the root system (and is an element of Weyl group of $\Phi$).

For each root $\alpha\in\Phi$, fix a non-zero $x_{\alpha}\in L_{\alpha}$ and $y_{\alpha}\in L_{-\alpha}$ so that $x_{\alpha}, y_{\alpha}, [x_{\alpha}, y_{\alpha}]$ span a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{C})$. Further $ad_{x_{\alpha}}$ and $ad_{y_{\alpha}}$ are nilpotent endomorphisms of $L$.

Humphreys' in his book on Lie algebra asserts that

If $\tau_{\alpha}:=\exp(ad_{x_{\alpha}}) \circ \exp(ad_{-y_{\alpha}}) \circ \exp(ad_{x_{\alpha}})$ then $\tau_{\alpha}$ is an automorphism of $L$ (clear!) which takes $L_{\beta}$ to $L_{\sigma_{\alpha}(\beta)}$.

Q. How to prove last statement in assertion (i.e. $\tau_{\alpha}(L_{\beta})=L_{\sigma_{\alpha}(\beta)}$?)

I initially tries taking $x_{\beta}\in L_{\beta}$ and see effect of $\tau_{\alpha}$ on it. But three automorphisms involved in $\tau_{\alpha}$ make complications. What other way I did is I tried to evaluate effect of $\tau_{\alpha}$ on $h_{\beta}$ where $h_{\beta}=[x_{\beta}, y_{\beta}]$ for $x_{\beta}\in L_{\beta}$, $y_{\beta}\in L_{-\beta}$ such that $span\{ x_{\beta}, y_{\beta}, h_{\beta}\}\cong \mathfrak{sl}(2,\mathbb{C})$.

Then an easy computation shows that $$\tau_{\alpha}(h_{\beta})=h_{\beta}-\alpha(h_{\beta})h_{\alpha}.$$ Beyond this I could not proceed towards proof of assertion. How to proceed?

What I just saw in Fulton-Harris is that he says,

it is sufficient to show that $\tau_{\alpha}(h)=h-\alpha(h)h_{\alpha}$ for all $h\in H$.

This almost completes a solution to my question; but something little I didn't understand. It is the following

Q. Why it is sufficient to show that $\tau_{\alpha}(h)=h-\alpha(h)h_{\alpha}$ for realizing $\tau_{\alpha}$ as automorphism of Lie algebra being induced by automorphism $\sigma_{\alpha}$ of root system?

  • $\begingroup$ As in your last question, why do you write a disjoint union in the Cartan decomposition? It's a direct sum, $H\oplus \bigoplus_{\alpha\in \Phi}L_\alpha$. (Sorry for nitpicking, the questions are good otherwise.) $\endgroup$ Commented Apr 28, 2018 at 7:32
  • $\begingroup$ This is the notation in Humphreys' Lie algebra. $\endgroup$
    – Beginner
    Commented Apr 28, 2018 at 7:35
  • $\begingroup$ Really? That surprises me a lot. (Also, "Cartan decomposition" usually refers to something else, whereas this would often just be called the "root space decomposition".) $\endgroup$ Commented Apr 28, 2018 at 7:38
  • $\begingroup$ yes; see page 35, last 3-4 lines. $\endgroup$
    – Beginner
    Commented Apr 28, 2018 at 7:40

1 Answer 1


The key point is that $\tau_\alpha$ is a Lie algebra automorphism of $L$. (For the interested: see section 2.3. of Humphreys.).

Let $x_\beta \in L_\beta$ be given, and let $h \in H$ be arbitrary. The formula $\tau_{\alpha}(h)=h-\alpha(h)h_{\alpha}$ leads to the following computation:

\begin{align}[h, \tau_\alpha(x_\beta)] &=[\tau_\alpha(h), \tau_\alpha(x_\beta)] +\alpha(h)[h_\alpha, \tau_\alpha(x_\beta)] \\ &=[\tau_\alpha(h), \tau_\alpha(x_\beta)] -\alpha(h)[\tau_\alpha(h_\alpha), \tau_\alpha(x_\beta)] \\&=\tau_\alpha[h,x_\beta]-\alpha(h)\tau_\alpha[h_\alpha, x_\beta] \\ &=\left( \beta(h)-\alpha(h)\beta(h_\alpha)\right)\tau_\alpha(x_\beta) \\&= \left( \beta-\langle\beta, \alpha \rangle \alpha \right)(h)\tau_\alpha(x_\beta) \\&= \sigma_\alpha(\beta)(h) \tau_\alpha (x_\beta)\end{align} This shows that $\tau_\alpha(x_\beta) \in L_{\sigma_\alpha(\beta)} $. That is, $\tau_{\alpha}(L_{\beta}) \subset L_{\sigma_{\alpha}(\beta)}$. Equality follows from the dimension argument.


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