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From Lie's theorem we know that the complex representations of any solvable Lie algebra $\mathfrak{g}$ over a subfield of $\Bbb{C}$ are such that there exists a basis in which the representation matrices are upper triangular. Nilpotent Lie algebras are solvable, so Lie's theorem applies to them as well. Algebras of upper diagonal (finite-dimensional) matrices with zero diagonal entries are nilpotent Lie algebras (with the commutator bracket); moreover, from Engel's theorem, we know that the adjoint representation of a nilpotent Lie algebra admits a basis in which the representation matrices are upper triangular with zero diagonal entries. But is it true that for the representations of any nilpotent Lie algebra over a subfield of $\Bbb{C}$ there exists a basis in which the representation matrices are upper triangular with zero diagonal entries complex representations?

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  • $\begingroup$ No. $\mathfrak{g} = \mathbb{R}$ is already a counterexample. $\endgroup$ – Qiaochu Yuan Feb 20 '16 at 4:38
  • $\begingroup$ Right. What about nilpotent (or even solvable) Lie algebras with zero center? $\endgroup$ – Giorgio Comitini Feb 20 '16 at 12:30
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    $\begingroup$ Nontrivial nilpotent Lie algebras always have nonzero center, see here. $\endgroup$ – Dietrich Burde Feb 20 '16 at 12:33
  • $\begingroup$ Ok. Then is there any hypothesis (apart from the trivial one: the algebra is an algebra of nilpotent endomorphisms) under which the representations have no diagonal entries)? $\endgroup$ – Giorgio Comitini Feb 20 '16 at 12:50
  • $\begingroup$ @Giorgio: no Lie algebra has this property. By Ado's theorem it must be nilpotent, so in particular solvable, so has nonzero abelianization. But a nonzero abelian Lie algebra has nontrivial 1-dimensional representations that cannot be strictly upper triangularized. $\endgroup$ – Qiaochu Yuan Feb 20 '16 at 16:20

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