What is an example of a real solvable simply-connected Lie group $G$ whose nilradical does not have a complement (that is, $G$ is not a semidirect product of the nilradical and another subgroup)? Is it possible to produce such a group if $G$ is supersolvable (that is, $G$ can be embedded in a group of upper triangular matrices)? Or the nilradical of a triangular real Lie group always split?


I know that such examples cannot be isomorphic to an algebraic group, because by the Jordan-Chevalley decomposition the unipotent radical (and therefore the nilradical) has a complement.



Example of a supersolvable Lie group/algebra whose nilradical does not have a complement
$\newcommand{\mk}{\mathfrak}$Let $\mk{h}\_{2n+1}$ be the $(2n+1)$-dimensional Lie algebra (basis $x\_1,\dotsc,x\_n,y\_1,\dotsc,y\_n,z$, nonzero brackets $[x\_i,y\_i]=z$).


The $n$-dimensional abelian Lie algebra $\mk{a}\_n$ (basis $a\_1,\dotsc,a\_n$) acts on it by $a\_i\cdot x\_i=x\_i$, $a\_i\cdot y\_i=-y\_i$, rest of the action being zero.


Through the quotient map $\mk{h}\_{2n+1}\to\mk{a}\_{2n}$ one thus gets an action of $\mk{h}\_{2n+1}$ on $\mk{h}\_{4n+1}$.


Then define the semidirect product $\mk{h}\_{2n+1}\ltimes\mk{h}\_{4n+1}$, and identify the central basis elements of the two.


This thus defines a $(6n+1)$-dimensional Lie algebra with basis $x\_i,y\_i,t\_i,u\_i,v\_i,w\_i,z$ ($1\le i\le n$) and nonzero brackets
\begin{gather\*}
[x\_i,t\_i]=t\_i,\quad [x\_i,v\_i]=-v\_i,\quad [y\_i,u\_i]=u\_i, \quad[y\_i,w\_i]=-w\_i, \\
[x\_i,y\_i]=[t\_i,v\_i]=[u\_i,w\_i]=z.
\end{gather\*}


This is supersolvable. The nilpotent radical has basis $t\_i,u\_i,v\_i,w\_i,z$ ($1\le i\le n$), so the quotient is abelian with basis $x\_i,y\_i$ ($1\le i\le n$). Observe that the isomorphic image of $\mk{h}\_{2n+1}$, i.e., with basis $x\_i,y\_i,z$ ($1\le i\le n$) is a Cartan subalgebra (= nilpotent, equal to its normalizer).


This does not split (for $n\ge 1$). Indeed, suppose by contradiction it does: let $\mk{a}$ be a splitting subalgebra. Then the subspace generated by $\mk{a}$ and $z$ is an abelian $(2n+1)$-dimensional subalgebra and is a Cartan subalgebra as well, unlike the previous one which is abelian. But in a solvable Lie algebra all Cartan subalgebras are conjugate (cf. Bourbaki) under automorphisms of the Lie algebra. This is a contradiction (for $n=1$ it can be directly checked: indeed $[x\_1+a,y\_1+b]$ is never zero for any $a,b$ in the nilpotent radical).




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Edit 1. Wondering if the above 7-dimensional example is of minimal dimension: ~~I check that the minimal dimension of an example is 6.~~ Here's ~~the~~ a 6-dimensional example (it depends in a nonzero parameter $\lambda$), without details (which are analogous to the previous one): basis $(x,y,t,u,v,w)$, nonzero brackets


$$[x,t]=\lambda t,\quad [x,u]=u,\quad [x,v]=-v,\quad [u,v]=[x,y]=z.$$


[By the next edit, this is not the smallest example. I think this becomes the smallest example if we require in addition that $[\mk{g},\operatorname{nil}(\mk{g})]=\operatorname{nil}(\mk{g})$, where $\operatorname{nil}(\mk{g})$ means the nilpotent radical.]




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Edit 2. Looking for a reference I found Bourbaki, Lie I, §5, exercise 6:


Consider the 5-dimensional Lie algebra with basis $(u,v,w,x,y)$ and nonzero brackets
$$[u,v]=w,\quad [u,x]=x,\quad[v,y]=y.$$


Check that the subspace with basis $(w,x,y)$ is the nilpotent radical, and that $\mathfrak{g}$ is not the semidirect product of an abelian subalgebra with a nilpotent ideal (hence the nilpotent radical doesn't split, since the quotient is abelian).


