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Let $L$ be a (finite-dimensional) complex semisimple Lie algebra. Then we know that $L = [L,L]$. Is it true that every element of $L$ must be a commutator?

Since a complex semisimple Lie algebra is a direct sum of simple Lie algebras, this question reduces to the case where $L$ is simple.

For example, we know that every complex matrix with trace zero is a commutator, see this question. So this is true when $L = \mathfrak{sl}_n(\mathbb{C})$.

But what about other families of simple Lie algebras? Is this known for classical simple Lie algebras? What about the exceptional Lie algebras?

Xander Henderson
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spin
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2 Answers2

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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.

Nazih Nahlus
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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.

Dietrich Burde
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    Thanks. So according to the paper, this holds over any infinite field. – spin Apr 28 '14 at 06:14
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    You are welcome. Yes, it even holds over finite fields of cardinality big enough, see the details in the proof of theorem $A$. – Dietrich Burde Apr 28 '14 at 07:47
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    Do I understand correctly though that it is always assumed in this paper that the Lie algebra is split? That's how I understand the first few lines of paragraph 2 in there. Also, otherwise this result would contradict statement 4) of the other answer, and make statement 2) redundant. – Torsten Schoeneberg Jan 04 '19 at 21:48