Consider a finite dimensional complex semi-simple Lie algebra $L$ with Cartan subalgebra $H$ (i.e. every $h\in H$ is $ad$-nilpotent). Denote $\Phi=\Phi(L,H)$ the set of roots.

Assume $\Phi=\Phi_1\cup\Phi_2$ for non empty $\Phi_i$ and $(\alpha,\beta)=0$ for all $\alpha\in\Phi_1,\beta\in\Phi_2$. Here $(\alpha,\beta)=\kappa(H^\alpha,H^\beta)$, where $\kappa$ is the Killing form and $H^\alpha$ is the unique element such that $\kappa (H^\alpha,\cdot)=\alpha$ (analogue for $H^\beta$). Set, for $i=1,2$, $$L_i=\text{span}_{\mathbb{C}}\{H^{\alpha}:\alpha\in\Phi_i\}\oplus\bigoplus_{\alpha\in\Phi_i}L_\alpha.$$

Why do we have $L=L_1\oplus L_2$ as Lie algebras?


1 Answer 1


It is obvious that $L=L_1\oplus L_2$ as vector spaces. To check that it holds as Lie algebras, you need to show that $[L_1,L_2]=0$. But this follows immediately from the orthogonality assumption, and the fact that $\alpha+\beta\notin\Phi$ for all $\alpha\in\Phi_1$ and $\beta\in\Phi_2$.

  • $\begingroup$ Thank you! Why is this clear as vector spaces? That $[L_1,L_2]=0$ I can show but why os this enough in order to see that the equality holds as Lie algebras? $\endgroup$
    – user455297
    Jun 16, 2017 at 19:39
  • $\begingroup$ It is clear as vector spaces because you have the Cartan decomposition $L=H\oplus\bigoplus_{\alpha\in\Phi}L_\alpha$. If $x=x_1+x_2\in L$ and $y=y_1+y_2\in L$ with $x_1,y_1\in L_1$ and $x_2,y_2\in L_2$, then you have $[x,y]=[x_1,y_1]+[x_2,y_2]$ because the terms $[x_1,y_2]$ and $[x_2,y_1]$ vanish. This is all you need to see that you have a direct sum as Lie algebras. $\endgroup$ Jun 17, 2017 at 11:34
  • $\begingroup$ Alright about the 2nd part, but the first one is still not clear for me. Why is the sum of the two spans direct and equal to $H$? $\endgroup$
    – user455297
    Jun 17, 2017 at 19:41
  • $\begingroup$ I think you can decompose $H$ as follows: You choose a set of simple roots $\Delta$ in $\Phi$. This set will decompose into a disjoint union $\Delta=\Delta_1\cup\Delta_2$ where $\Delta_i$ are simple roots in $\Phi_i$. Now, $H_\alpha$ where $\alpha\in\Delta$ form a basis of $H$, and similar for the two spans, you have bases indexed by $\Delta_i$. The direct sum decomposition of $H$ just corresponds to splitting the basis according to the dijoint union of $\Delta$. $\endgroup$ Jun 18, 2017 at 6:53

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