I have some elementary questions about Lie algebras and vector space complements.

Let $(\mathfrak{g},[.,.])$ be a finite-dimensional Lie algebra and $\mathfrak{g}_1$ a Lie ideal in $\mathfrak{g}$.

1) Is it possible to choose $\mathfrak{g}$ and $\mathfrak{g}_1$ in such a way that there is no vector space complement $\mathfrak{g}_2$ of $\mathfrak{g}_1$ in $\mathfrak{g}$ which is additionally a sub-Lie algebra of $\mathfrak{g}$?

2) Is there an example of $\mathfrak{g}$ and $\mathfrak{g}_1$ such that there is a vector space complement $\mathfrak{g}_2$ of $\mathfrak{g}_1$ in $\mathfrak{g}$ with $[\xi_1,\xi_2] = 0$ for all $\xi_1 \in \mathfrak{g}_1$ and $\xi_2 \in \mathfrak{g}_2$ but such that no such complement is a Lie ideal in $\mathfrak{g}$?

In the situation I need, all Lie algebras occur as Lie algebras of some Lie group. For every Lie algebra $\mathfrak{g}$ there exists a Lie group $G$ with $\mathrm{Lie}(G) = \mathfrak{g}$, but that's rather abstract, so it would be nice if $\mathfrak{g}$ and $\mathfrak{g}_1$ were the Lie algebras of some well-known Lie groups.

I'm not an expert in Lie algebra theory, so if some things described in 1) or 2) have special names in the literature it would be nice if you could mention that or give some references.

  • $\begingroup$ For this kind of question, it's important to specify the base field and whether it is algebraically closed. $\endgroup$ – Jim Humphreys Jul 13 '10 at 11:45
  • $\begingroup$ As a base field I want the real numbers or the complex numbers. $\endgroup$ – student Jul 13 '10 at 12:33
  • 1
    $\begingroup$ @Jim: JFF's answer below is not (very) dependent on the choice of field. $\endgroup$ – Theo Johnson-Freyd Jul 13 '10 at 16:56
  • $\begingroup$ Are there special names in the literature for the situations in 1) and 2) I mentioned ? I have found something about 2), if the lie-Algebras arise as Lie-Algebras of some connected Lie-Groups, $G$ resp. $G_1$ ($G_1$ normal Lie-subgroup in $G$), having a vector space complement $\mathfrak{g}_2$ with $[\mathfrak{g}_1,\mathfrak{g}_2] = 0$ means that $\mathfrak{g}_2$ ist $G_1$-invariant. And such a homogenous space $G/G_1$ is called reductive, I have found that in O'Neill, Semi-Riemannian-Geometry and Baums "Eichfeldtheorie" book. But I havn't found any more comprehensive references. $\endgroup$ – student Jul 16 '10 at 13:42

Take $\mathfrak{g}$ to be the Lie algebra of the Heisenberg Lie group (i.e., the nilpotent Lie algebra of strictly upper triangular $3\times 3$ matrices) and let $\mathfrak{g}_1$ be the centre. This gives you examples to (1) and (2).


Now that I have more time, I can perhaps say something else about (1). (I will work in zero characteristic, just in case.)

Every ideal $\mathfrak{g}_1$ of $\mathfrak{g}$ gives rise to an exact sequence of Lie algebras $$\begin{matrix} 0 & \longrightarrow & \mathfrak{g}_1 & \longrightarrow & \mathfrak{g} & \longrightarrow & \mathfrak{h} & \longrightarrow & 0 \end{matrix}$$ where $\mathfrak{h} = \mathfrak{g}/\mathfrak{g}_1$. This says that $\mathfrak{g}$ is an extension of $\mathfrak{h}$ by $\mathfrak{g}_1$.

Your condition (1), that there should not be a Lie subalgebra of $\mathfrak{g}$ complementary to $\mathfrak{g}_1$, is precisely the statement that the sequence does not split. Hence, in particular, $\mathfrak{g}$ is not isomorphic to a (semi)direct product of $\mathfrak{g}_1$ and $\mathfrak{h}$.

If $\mathfrak{g}_1$ were abelian, it becomes in a natural way an $\mathfrak{h}$-module and you can classify such extensions by the Lie algebra cohomology group $H^2(\mathfrak{h},\mathfrak{g}_1)$. Then your condition (1) can be rephrased as saying that the corresponding extension class in $H^2(\mathfrak{h},\mathfrak{g}_1)$ is nonzero. (I do not know what classifies extensions in the general case.)

  • $\begingroup$ Thanks! Do you have some references for what you added? $\endgroup$ – student Jul 16 '10 at 13:32
  • $\begingroup$ I learned this from Hilton and Stammbach's A course in homological algebra: books.google.co.uk/books?id=2VNrHuweuokC Probably there are more modern references and perhaps other people might chime in. $\endgroup$ – José Figueroa-O'Farrill Jul 16 '10 at 23:59

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