Copies of ax+b inside the AN part of an Iwasawa decomposition?

Question

As a relative novice to the structure theory of Lie algebras and Lie groups, the following is what I can gather from reading parts of Helgason's book DG, Lie groups and symmetric spaces and Knapp's Beyond an introduction. $\newcommand{\Real}{{\mathbb R}}\newcommand{\fa}{{\sf a}}\newcommand{\fg}{{\sf g}}\newcommand{\fk}{{\sf k}}\newcommand{\fn}{{\sf n}}$

Let $G$ be a connected, semisimple, non-compact Lie group (over the reals, not necessarily a matrix group, and not necessarily with finite centre), let $\fg=\fk\oplus \fa\oplus\fn$ be an Iwasawa decomposition of its Lie algebra, and let $G=KAN$ be the corresponding "global" Iwasawa decomposition. Since $G$ is non-compact, both $A$ and $N$ are non-trivial; moreover, both are in fact closed, simply connected subgroups of $G$.

It is observed in Helgason's book that if $\fg$ is the Lie algebra of $G$ and $\fn$ is the nilpotent subalgebra constructed in the Iwasawa decomposition at the Lie algebra level, then for each $x\in \fn$ we can find $h\in\fg$ such that $[h,x]=x$. (This is one of the steps in the proof of the Jacobson-Morozov theorem on existence of copies of ${\sf sl}(2,\Real)$ inside $\fg$.)

Q1. Can we always find $x\in\fn$, $x\neq 0$, and $h\in\fa$, such that $[h,x]=x$?

I suspect that one could modify Jacobson's argument (Proc. AMS 1951) or the version Helgason gives, due to Kostant (Amer. J. Math. 1959, Section 3) to do this, but I have not succeeded in nailing down an argument.

If the answer to Q1 is positive, then I think we should be able to exponentiate $h$ and $x$ and obtain a closed and simply connected subgroup of $AN$, which would then be isomorphic to the so-called $ax+b$ group ${\Real}\rtimes {\Real}_+^*$. (Of course one can see this directly for $SL(2,\Real)$ and thence for $SL(n,\Real)$, which is the case that motivated the present question.)

Q2. Assuming it is true that $AN$ always contains a closed copy of the $ax+b$ group, is this standard knowledge for which there exists a reasonably precise citation?

Answer 1

Here is an elementary argument addressing Q1. The adjoint action of the subalgebra $\mathfrak{a}$ on $\mathfrak{g}$ has the following two key properties:

1. It is diagonalizable (the operators $ad(H)$ for $H\in\mathfrak{a}$ are symmetric with respect to the Killing form);

2. It stabilizes $\mathfrak{n}$.

It follows that either $\mathfrak{a}$ centralizes $\mathfrak{n}$ or there exists an element $H\in\mathfrak{a}$ with a non-zero eigenvalue. In the latter case, rescaling, we get a pair $H\in\mathfrak{a}, X\in\mathfrak{n}$ such that $[H,X]=ad(H)X=X.$ As Yves has already remarked, the former case is never realized since the Lie algebra $\mathfrak{g}$ is semisimple.

Answer 2

Every linear algebraic group over a field $K$ of characteristic zero that is not nilpotent-by-anisotropic contains a Zariski closed subgroup isomorphic over $K$ to a non-direct semidirect product of the multiplicative group by the additive group. In particular the real points contain an $ax+b$-group. This can be applied to $AN$.

Proof: if the reductive quotient is not abelian-by-anisotropic, then it contains a subgroup isogeneous to $\text{SL}_2$ and we're done. Otherwise, the connected isotropic part $D$ of the reductive Levi factor (which is abelian) does not centralize the unipotent radical. Define weights on the Lie algebra $\mathfrak{u}$ according to the action of $D$. Pick a nonzero weight $\alpha$ and consider the corresponding subgroup $\mathfrak{u}_\alpha$. Changing $\alpha$ if necessary, we can suppose that $2\alpha$ is not a weight, so $\mathfrak{u}_\alpha$ is commutative. Then it is easy to find a group as required in $DU_\alpha$.

[In your setting, since $A$ is an isotropic connected torus and $N$ is unipotent, to say that $AN$ is not nilpotent-by-anisotropic just means that $A$ does not centralize $N$; there are too many ways to check this to pick one :)]