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Let $\frak{g}$ be a semi-simple Lie algebra of finite dimension and let $\frak{h}, \frak{m}$ be ideals of $\frak{g}$. If $k$ denotes the Cartan Killing form, prove that the following statements are equivalent.

(i) $\frak{h}\cap\frak{m} = \lbrace{0}\rbrace$

(ii) $[\frak{h},\frak{m}] = \lbrace{0}\rbrace$

(iii) $k(h,m)=0\,\,$ for all $\,\,h\in\frak{h}\,\,$ and $\,\,m\in\frak{m}$.

In my proof, (i) implies (ii) follows that $[\frak{h},\frak{m}]\subset \frak{h}\cap\frak{m}$ and (ii) implies (iii) it follows that $\frak{h}$ and $\frak{m}$ are semi-simples and of the associativity of the Cartan Killing form.

Could someone give me a suggestion for part (iii) implies (i)?

José Carlos Santos
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José Sanchez
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1 Answers1

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I assume you have the "Cartan criterion", which says that on a semisimple Lie algebra, the Killing form is non-degenerate (the converse holds as well, but we don't need that).

Now to show (iii) implies (i), what what would $k(x,x)$ be (according to (iii)) for every $x \in \mathfrak{h} \cap \mathfrak{m}$? So by the above criterion ...

Torsten Schoeneberg
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