The intersection of all Levi factors

Question

Let $\mathfrak g$ be a Lie algebra over $\mathbb R$ and let $\mathbb s$ be a Levi factor, i.e, a maximal semisimple Lie subalgebra. Since Levi factors are conjugate to each other, what is the intersection of all these conjugation subalgebras? i.e, what is the intersection of all Levi factors? Can it be trivial?

Answer 1

Since two Levi factors are always conjugate by an inner automorphism, the intersection of all conjugates of a Levi factor $\mathfrak q$ is equal to $$ \mathfrak h =\bigcap_{g\in G}\operatorname{Ad}_g(\mathfrak q) $$ Therefore, $\mathfrak h$ is invariant under all one-parameter subgroups, so by differentiating $\operatorname{Ad}$ we see that it is an ideal in $\mathfrak g$.

Now, we claim that this ideal is exactly the centraliser of the solvable radical $\operatorname{rad}\mathfrak g$ in $\mathfrak q$.

Indeed, clearly $\mathfrak h\cap \operatorname{rad}\mathfrak g = \{0\}$, so for $h\in\mathfrak h$, $x\in\operatorname{rad}\mathfrak g$ we get $[h,x]=0$.

On the other hand, by a result of Mal'cev two Levi factors are conjugate by an inner automorphism coming from the nilradical (so in the above formula it's enough to conjugate by $g= \exp(x)$ for $x\in\operatorname{nil}\mathfrak g$). Therefore, if $h$ commutes with the radical, it's acted trivially upon by the nilradical and hence is in $\mathfrak h$.