1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $\beta\in\Phi_2$.

$\endgroup$
  • $\begingroup$ 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 '17 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. $\endgroup$ – user213008 Jun 17 '17 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 '17 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 $\Delta$. $\endgroup$ – user213008 Jun 18 '17 at 6:53

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.