Let $\mathfrak{g}$ be a complex linear Lie algebra of dimension $n$. If there exists a basis $\{e_1,\dots,e_n\}$ of $\mathfrak{g}$ such that $\begin{equation}\sum_{i=1}^n[e_i,\bar{e_i}^T]=0,\end{equation}$ what can we say about $\mathfrak{g}$? Is it true that every semisimple $\mathfrak{g}$ has this property? Thanks!
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$\begingroup$ What is $\bar{e_i}^T$? $\endgroup$– abxCommented Jul 25, 2014 at 8:43
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$\begingroup$ @abx: That is the conjugate transpose of $e_i$. $\endgroup$– PiojoCommented Jul 25, 2014 at 8:45
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2$\begingroup$ “conjugate transpose” is not something that makes sense in an abstract Lie algebra. So I suppose you are implicitly assuming that $\mathfrak{g}$ is a complex Lie subalgebra of $\mathfrak{gl}(N,\mathbb{C})$ for some $N$? $\endgroup$– José Figueroa-O'FarrillCommented Jul 25, 2014 at 12:45
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1$\begingroup$ @JoséFigueroa-O'Farrill: that is exactly what I mean by a linear Lie algebra. $\endgroup$– PiojoCommented Jul 25, 2014 at 12:54
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$\begingroup$ My apologies. I read “complex linear” as “complex-linear”, meaning simply “complex”. Now I understand the question; but have to think about an answer. $\endgroup$– José Figueroa-O'FarrillCommented Jul 25, 2014 at 20:33
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This condition seems awkward from a Lie algebra point of view. Perhaps you could explain where it arises from?
A few observations: -A sufficient condition is to have a basis of normal matrices. I believe one can construct a basis of normal matrices for the classical complex lie algebras. I am not sure about the exceptional lie algebras.
-This condition is closed under direct sums, so you can use this to build semisimple lie algebras
-A (silly) example of a lie algebra where this condition doesn't hold is just the one dimensional lie algebra spanned by the matrix $E = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$ since $E E^* - E^* E = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}$.
