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Given a Lie algebra (finite-dimensional, over a field) and a basis, denote the structure constants in the usual way: $[e_i,e_j]=\sum_kc_{ij}^ke_k$. We say that the structure constants are cyclic if $c_{ij}^k=c_{jk}^i$ for all $i,j,k$ (note that this depends on the choice of basis).

The Lie algebra being given, the existence of a basis with cyclic structure constant is a nontrivial condition as it can easily be checked not to exist in a non-abelian 2-dimensional Lie algebra.

Which Lie algebras admit a basis with cyclic structure constants?

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    $\begingroup$ Why does $c^k_{ij} = c^i_{jk}$ define the semisimples? In $\mathfrak{sl}_2$, the bracket $\left[e,f\right]$ has a nonzero $h$-coordinate, but the bracket $\left[h,e\right]$ has no nonzero $f$-coordinate. Are you using a weirder basis? $\endgroup$
    – darij grinberg
    Commented Oct 12, 2016 at 20:55
  • $\begingroup$ Typo. Sorry. Corrected. $\endgroup$
    – Hauke Reddmann
    Commented Oct 14, 2016 at 18:25
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    $\begingroup$ Endow your Lie algebra with the symmetric bilinear form making $\{e_i\}$ an orthornormal basis (the Gram matrix is the identity matrix). Since $c_{ij}^k=([e_i,e_j],e_k)$, what you call "cyclicity" is equivalent to the invariance of the form. $\endgroup$
    – Victor Protsak
    Commented Oct 14, 2016 at 20:55
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    $\begingroup$ I rewrote the question hoping that it will it be reopened so that Victor's answer can be posted. $\endgroup$
    – YCor
    Commented Oct 15, 2016 at 5:20
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    $\begingroup$ @Protsak: does it mean that every semi-simple Lie algebra over char zero field has cyclic structure constants? (Because the Killing form is symmetric and non-degenerate) $\endgroup$
    – Venkataramana
    Commented Oct 15, 2016 at 8:35

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Endow your algebra with the symmetric bilinear form making $\{e_i\}$ an orthornormal basis (the Gram matrix is the identity matrix). Since $c_{ij}^k=([e_i,e_j],e_k)$, the cyclicity condition is equivalent to the invariance of the form: $$ ([e_i,e_j],e_k)=(e_i,[e_j,e_k]). $$ Thus a finite-dimensional Lie algebra admits a basis with cyclic structure constants if and only if it admits an invariant symmetric bilinear form with Gram matrix the identity matrix. If the base field is algebraically closed not of characteristic 2, the latter condition simplifies to "admits a nondegenerate invariant symmetric bilinear form".

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  • $\begingroup$ Also in the real case, this exists if and only if the Lie algebra is reductive compact. Interestingly, in the complex case, the class of such Lie algebras is much larger since they include non-reductive ones. $\endgroup$
    – YCor
    Commented Oct 15, 2016 at 14:10
  • $\begingroup$ Yes, they are known as "quadratic Lie algebras". $\endgroup$
    – Victor Protsak
    Commented Oct 15, 2016 at 15:30
  • $\begingroup$ They (Lie algebras endowed with an invariant symmetric bilinear form, or which admit at least such a form) have a multitude of names, which I tried to gather in Remark 2.5 in normalesup.org/~cornulier/Koszul.pdf. $\endgroup$
    – YCor
    Commented Oct 15, 2016 at 16:43
  • $\begingroup$ Mind to give a small dimension example? (Non-semisimple, of course. Over the complex numbers is OK.) $\endgroup$
    – Hauke Reddmann
    Commented Oct 16, 2016 at 22:22

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