# Decomposition of Lie algebra: do the simple and maximal torus parts commute?

I have the following exercise:

Consider a Lie algebra $$\mathfrak{g}$$. Decompose $$\mathfrak{g}$$ using the Levi decomposition, so $$\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{r}$$. Let $$\mathfrak{a}$$ be a maximal torus inside the radical. Is it always true that $$[\mathfrak{s},\mathfrak{a}]=0$$? Discuss which hypotesis are needed to obtain such a result.

Now I started assuming that $$\mathfrak{g}$$ is algebraic (to avoid exceptional or strange cases, if any), and I procede as suggested, so I get $$\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{a}\oplus\mathfrak{m}$$ for some complement $$\mathfrak{m}$$ of $$\mathfrak{a}$$ in the radical $$\mathfrak{r}$$. For sure $$[\mathfrak{s},\mathfrak{a}]\subset \mathfrak{r}$$. But then I don't have any idea, nor I can prove it is true always (I guess it is false).

For the second question, I can think only of "trivial" answers: it is true for solvable Lie algebras and for semi-simple Lie algebras. For characterisitic zero, it is true for reductive Lie algebra, because the radical is the center, so it is abelian and commute with everythings.

Suggestions?

• Sorry for a maybe dumb question, but the exact definition for "a maximal torus inside the radical" is ...? Commented May 4, 2021 at 19:47
• Beside that, have you looked at standard non-trivial examples like standard parabolic subalgebras of $\mathfrak{gl}_n$? Commented May 4, 2021 at 19:48
• @TorstenSchoeneberg from my notes, a maximal torus inside the radical is an abelian subalgebra of the radical that is not contained in any other abelian subalgebras of the radical. I didn't check the Parabolic, yet. Commented May 4, 2021 at 21:43

Hint: Any element $$x$$ of a Lie algebra (over a field $$k$$ say) is contained in the abelian Lie subalgebra $$k \cdot x$$. Hence by your definition of maximal torus, by a "finite dimension" argument, every element of $$\mathfrak r$$ is contained in some maximal torus $$\mathfrak a$$ inside $$\mathfrak r$$.
That means that for $$\mathfrak s$$ to commute with all such maximal tori $$\mathfrak a$$, it has to commute with ...
• Got it, so the "trick" is that the property is for any maximall torus. Hence $\mathfrak{s}$ commute with all $x\in\mathfrak{r}$, hence the radical is the center. Am I right? Thank you Commented May 5, 2021 at 7:15
• Well that's how I read the question (because then it has an easy answer). If we're just asked how to tell whether one given torus in the radical commutes with the semisimple part, it's harder. In general, probably, some tori in the radical commute with $\mathfrak s$ and some don't, and I wouldn't see right away a non-tautological criterion to tell them apart. Commented May 5, 2021 at 14:57