Reductive space & Reductive Lie algebra

If $M=G/H$ is a reductive space and $\mathfrak{g}=\mathfrak{h}+\mathfrak{m}$ be the canonical decomposition, then are $\mathfrak{g}$ or $\mathfrak{h}$ or both reductive lie algebras? (in this case, where can I find a proof?)

It is for every ideal $\mathfrak{a}$ of $\mathfrak{g}$ there exists an ideal $\mathfrak{b}$ sucht that $\mathfrak{g}=\mathfrak{a}+\mathfrak{b}$ or equivalently the adjoint representation is semi-simple

• What is a reductive space? – Mikhail Borovoi Feb 23 '15 at 4:10
• @MikhailBorovoi According to encyclopedias it means that $\mathfrak h$ $H$-equivariantly splits out of $\mathfrak g$. – მამუკა ჯიბლაძე Feb 23 '15 at 8:45
• Also I think $G$ and $H$ must be reductive groups, with $H$ closed subgroup of $G$ (so that $G/H$ is a reductive space means $(G,H)$ is a reductive pair). – მამუკა ჯიბლაძე Feb 23 '15 at 8:51

It depends on the definition of a reductive space $G/H$. Some authors require that in addition to $\mathfrak{g}=\mathfrak{h}+\mathfrak{m}$, $[\mathfrak{h},\mathfrak{m}]\subseteq \mathfrak{m}$, we have that $(\mathfrak{g},\mathfrak{h})$ is a reductive pair of Lie algebras, by which they mean that $\mathfrak{g}$ is reductive, and $\mathfrak{h}$ is reductive in $\mathfrak{g}$. In this case, since $\mathfrak{h}$ is a $\mathfrak{h}$-submodule of $\mathfrak{g}$, this would imply that $\mathfrak{h}$ itself is reductive.
• but for example take $\mathfrak{g}=e_{1}+e_{2}$ as a vector space and the commutator $[e_{1},e_{2}]=e_{2}$ then $\mathfrak{g}$ is not a reductive lie algebra and take $\mathfrak{h}=e_{1}$ then $Lie(\mathfrak{g})/Lie(\mathfrak{h})$ Is a reductive space – user66943 Feb 23 '15 at 16:27
• Yes, it depends on the definition. Do you have a reference where your example $Lie(g)/Lie(h)$ is called a reductive space ? – Dietrich Burde Feb 24 '15 at 10:21
• @user66943: Your notation such as $Lie(\mathfrak{g})$ seems out of control here. And as Dietrich points out, precise definitions are needed. – Jim Humphreys Feb 24 '15 at 19:08