Is every element of a complex semisimple Lie algebra a commutator?

Solution 1:

Yes, this holds for all complex simple Lie algebras. A reference is Theorem $A$ in the article On commutators in a simple Lie algebra. The result can be extended to simple Lie algebras over more general fields.

Solution 2:

1) In addition to Brown Theorem above, it is worth noting the following "1.5" generators of such Lie algebras.

Theorem: Let L be a simple Lie algebra over an infinite field k of char 0 (or char not 2,or 3). Then L=[L, a] +[L, b] for some a, b in L. In particular, every element of L is a sum of at most 2 commutators.

Reference: G. Bergman, N. Nahlus, Homomorphisms of infinite direct product algebras especially Lie algebras, J. Algebra, 2011, pp. 67-104. (Thm. 26)

In fact, there are some explicit "1.5" generators over C (complex field) which can be found in some papers by Panyushev.

2) Back to "every element of L must be a single commutator", It is indeed true in COMPACT real semisimple Lie algebras

References: 1) For a proof using Kostant Convexity Theorem, see Appendix 3 in the book by Karl Hofmann & Sidney Morris, The Lie Theory of connected Pro-Lie groups, 2007. In fact, this proof was communicated to the authors by Karl-Hermann Neeb.

2) For another "implicit" proof, see Thm. 3.4 in "Some Questions about Semisimple Lie Groups Originating in Matrix Theory", Canad. J. Math. Vol. 46 (3), 2003, pp. 332-343 by Dragomir Z. Dokovi´c and Tin-Yau Tam. Their proof uses Dynkin diagrams

3) Another reference is by D. Akhiezer, On the Commutator Map for Real Semisimple Lie Algebras, 2015, arXiv:1501.02934

4) The question whether "every element of L must be a single commutator for any Real semisimple Lie algebra L" is an open question.