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Let $\mathfrak{g}$ be a Frobenius Lie algebra, that is, there exist $f \in \mathfrak{g}^{*}$ such that the bilinear form defined by $b(x,y)=f([x,y])$ is non-degenerate. Since $b$ is non-degenerate there exist a unique $x_p \in \mathfrak{g}$ called the principal element(asociated to $f$) which satisfies $$ f \circ ad(x_p)=f $$ and it can be proved that $$tr(ad(x_p))=\frac{\dim (\mathfrak{g})}{2}$$ which implies that $x_p \notin [\mathfrak{g},\mathfrak{g}]$. Here is my question: Why if $\lbrace x_p,x_1,\ldots, x_{2n+1} \rbrace$ is a basis of $\mathfrak{g}$ then $span\lbrace x_1,\ldots, x_{2n+1} \rbrace$ is an ideal of $\mathfrak{g}$?.

I read in an article, namely, Principal derivations and codimension one ideals in contact and Frobenius Lie algebras that this follows from the last remark(that is, $x_p \notin [\mathfrak{g},\mathfrak{g}]$) but I don't agree with this. I mean, if we consider the $2$-dimensional Lie algebra $\mathfrak{g}$ defined by $[e_1,e_2]=e_1+e_2$, then $e_1 \notin [\mathfrak{g},\mathfrak{g}]$ but $span \lbrace e_2 \rbrace$ is not an ideal of $\mathfrak{g}$, am I wrong? If so, why?

In advance, thank you.

ferolimen
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1 Answers1

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Your counterexample to the claim, which appears right before Theorem 3.1 in the article 1, looks correct to me.

What is true, and maybe was what the authors had in mind there , is that any subspace of a Lie algebra $L$ which contains $[L,L]$ is an ideal; but of course $x_p \notin [L,L]$ does not mean that just any vector space complement of the space $span\{x_p\}$ contains $[L, L]$, as your example also shows.

Actually, if $\{x_p, x_1, ..., x_{2n+1}\}$ is a vector space basis of $L$ and the span of $\{x_1, ..., x_{2n+1}\}$ is an ideal, it also follows that $[L,L]$ is contained in that ideal; in other words, $[L, L] \subseteq span\{x_1, ..., x_{2n+1}\}$ is a necessary and sufficient condition for that span to be an ideal. (And it is easy to come up with counterexamples like yours where that is not the case; indeed, as long as $L$ is not abelian i.e. $[L,L] \neq 0$, we can "shift" any basis for which it is an ideal to a slightly different one where it's not, e.g. replacing an arbitrary basis vector $x_i$ in $[L,L]$ by $x_p+x_i$.)

To fix the issue, one could insert the condition that $span\{x_1, .., x_{2n+1}\}$ contains $[L,L]$; or, explicitly choose a basis $\{x_1, ..., x_r\}$ of $[L,L]$ first, then extend the linearly independent set $\{x_p, x_1, ..., x_r\}$ to a basis $\{x_p, x_1, ..., x_{2n+1}\}$ of $L$.

1 T. Barajas, E. Roque & G. Salgado (2019) Principal derivations and codimension one ideals in contact and Frobenius Lie algebras, Communications in Algebra, 47:12, 5380-5391, DOI: 10.1080/00927872.2019.1623238

Torsten Schoeneberg
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