Existence of a real eigenvalue is a necessary condition for the density of all the orbits of a Lie subgroup of $GL(\mathbb{R},d)$

Question

Good morning,

I would like to pose the following (maybe naive) question. Let $\mathfrak{a}\subset \mathfrak{gl}(\mathbb{R},d)$ be any lie subalgebra, and $A$ be the connected, simply connected subgroup of $GL(\mathbb{R},d)$ generated by $\mathfrak{a}$.

Assume that for every $x\in\mathbb{R}^d\setminus \{0\}$, the orbit $Ax$ is dense in $\mathbb{R}^d$. Is it true that necessarily there has to be a matrix in $\mathfrak{a}$ which has a nonzero real eigenvalue?

Intuitively I am tempted to say so, since this eigenvalue would cover the ``radial'' part of any such orbit, but I have not been able to come across any similar results in the literature. I have looked for results in representation theory and in the theory of dynamical systems (I have looked for minimal actions of groups), but up to now I have found very scarce references.

Do you know any good reference to have a look at for such kind of questions? Or is my intuition just plainly false? Any counterexample? I'm a bit lost, so any help would be greatly appreciated.

Regards.

Answer 1

It's true.

Since $A$ acts irreducibly, so does $\mathfrak{a}$, so the latter is reductive. Write $\mathfrak{a}=\mathfrak{z}\oplus\mathfrak{s}$, with $\mathfrak{z}$ its center and $\mathfrak{s}=[\mathfrak{a},\mathfrak{a}]$ being semisimple.

Assume that $A$ has no element with a nonzero real eigenvalue. This implies that $\mathfrak{s}$ has real rank zero (so $[A,A]$ is compact semisimple).

If $\mathfrak{s} $is nonzero, it contains a nonzero element $s$. Necessarily $s$ has a nonzero eigenvalue (in $i\mathbf{R}$), say $i$ up to rescaling. Consider $z\in\mathfrak{z}$; since it centralizes an irreducible representation, we have a decomposition of $\mathbf{C}^d$ into two $z$-eigenspaces $V_1\oplus V_2$, on which the eigenvalues of $z$ are $a+ib$ and $a-ib$ for some $(a,b)\in\mathbf{R}^2$. Since $s$ commutes with $z$, the eigenspace of $s$ for the eigenvalue $i$ meets at least one of these two spaces. Therefore either $z+bs$ or $z-bs$ has the real eigenvalue $a$. If $a\neq 0$, we are done.

The argument works unless every element of $\mathfrak{z}$ has only eigenvalues in $i\mathbf{R}$, but in this case, $A$ has a compact closure and this is not compatible with the density assumption.

Finally, if $\mathfrak{s}=0$, then $A$ is abelian, so irreducibility implies that $d\in\{1,2\}$. If $d=1$ we are done. If $d=2$, then the abelian irreducible subalgebras are conjugate into the set of similarity matrices. If 2-dimensional, it contains real scalar matrices. Otherwise $A$ is 1-dimensional and cannot have dense orbits on $\mathbf{R}^2$.