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Let $\mathfrak{g}$ be a Lie subalgebra of the real upper triangular matrices (i.e. $\mathfrak{g}$ is a completely solvable real Lie algebra). $\mathfrak{g}$ has an exponential radical $\mathfrak{r}$, which is the intersection of the terms of its lower central series. Is it true that any element of $\mathfrak{g}\setminus \mathfrak{r}$ is contained in some Cartan subalgebra? Or more generally is there a necessary and sufficient condition on $\mathfrak{g}$ so that this holds?

The closest result we found is in Bourbaki (Groupes et algebres de Lie, chapitre 7, §3, Proposition 4), stating: let $\mathfrak{g}$ be a Lie algebra of rank $l$ over a field of characteristic $0$. Let $c$ be the nilpotency class of its Cartan subalgebras, and $X\in \mathfrak{g}$. There exists a subalgebra of dimension $l$ with nilpotency class at most $c$ containing $X$.

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    $\begingroup$ No. In the example of upper triangular matrices, elements of Cartan subalgebras are diagonalizable matrices. Elements of the exponential radical are nilpotent matrices. So, pick any (upper triangular) matrix that is neither nilpotent nor diagonalizable. $\endgroup$
    – YCor
    Commented Jul 9 at 8:32
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    $\begingroup$ Your last statement is not true. If $J$ is a Jordan block of maximal rank, its centralizer has dimension $n$ (the ambient dimension), which equals the rank, but this is not a Cartan subalgebra. $\endgroup$
    – YCor
    Commented Jul 9 at 8:33
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    $\begingroup$ It's not clear what you mean by a necessary and sufficient condition on $\newcommand\f{\mathfrak}\f g$ for your statement to hold, since $\f g$ has been fixed and, as @YCor points out, your statement is not true. However, perhaps you mean for $\f g$ now to be a general algebraic Lie algebra whose CSA are (Lie alg. of) tori, and $\f r$ its nilpotent radical. If $\f g/\f r$ is not a torus, then you can lift a non-semisimple element to $\f g \setminus \f r$; it will not belong in a CSA. If $\f g/\f r$ is a torus, then you need every ss element of $\f g \setminus \f r$ to act freely on $\f r$. $\endgroup$
    – LSpice
    Commented Jul 9 at 15:05
  • $\begingroup$ For example, the "$ax + b$ Lie algebra" $\operatorname{Add} \ltimes \operatorname{Add}$ has this property. If $s_1$ and $s_2$ are linearly independent, commuting, semisimple elements of $\mathfrak g \setminus \mathfrak r$, and $X$ is a simultaneous weight vector for them in $\mathfrak r$, then (because the weights are non-$0$) some linear combination of $s_1$ and $s_2$ annihilates $X$, which is a contradiction. So $\mathfrak g/\mathfrak r$ is $1$-dimensional, or $\mathfrak r$ is $0$. $\endgroup$
    – LSpice
    Commented Jul 9 at 15:22
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    $\begingroup$ @LSpice $\mathfrak{r}$ is usually much smaller than the nilpotent radical. For instance, if $\mathfrak{g}$ is nilpotent, then the exponential radical is zero (while the nilpotent radical is the whole algebra), and $\mathfrak{g}$ itself is its unique Cartan subalgebra. $\endgroup$
    – YCor
    Commented Jul 9 at 23:43

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