1

Let L be a Lie algebra and $\{e_i,h_i,f_i; i=1 \ldots l\}$ be a basis of L. Is it true that the subalgebra of L generated by $\{e_i,h_i,f_i; i=1 \ldots l\}$ equals whole of L? What is the form of the elements of subalgebra generated by $\{e_i,h_i,f_i; i=1 \ldots l\}$? I am not getting form a subalgebra generated by a finite subset of a Lie algebra L?

budi
  • 1,780
  • 2
    By definition, the Lie subalgebra of $L$ generated by a subset $S$ of $L$ is the least (wrt inclusion) Lie subalgebra of $L$ that contains $S$. If $S$ is a basis of (the vector space) $L$ then, since a Lie subalgebra must be a subspace of the underlying space, the Lie subalgebra of $L$ generated by $S$ will be $L$. – gniourf_gniourf Dec 18 '16 at 16:24
  • 1
    So what is the form of elements of a Lie subalgebra generated by a subset S of L where S need not be a basis. I think the elements are linear combinations of possible Lie brackets of elements of S. Am I right? – budi Dec 18 '16 at 16:27
  • 1
    Yeah, that's correct: linear combinations of all the possible Lie brackets of elements of $S$; all of them: not just $[e_1,e_2]$, $[e_1,e_3]$, etc. you also need all the nested ones, e.g., $[e_1,[e_1,[e_1,e_2]]]$, etc. – gniourf_gniourf Dec 18 '16 at 16:30
  • Yes yes I meant all of them. Thank you...... – budi Dec 18 '16 at 16:31

1 Answers1

1

The above set of generators is the classical set of generators of a complex simple Lie algebra of rank $l$. This is usually called the Chevalley generators' set, which satisfying some relations related to a Cartan matrix, you can find them for instance in H. Samelson - Notes on Lie algebras, p.73 or N. Jacobson - Lie algebras, p.125. The generating space of a set S is the sum $\sum_{i\geq0}S_i$, where $S_0=S$ and $S_i=[S_{0},S_{i-1}]$, $i\geq1$, which is the least subalgebra containing $S$.

Youness EL KHARRAF
  • 39