Is a toral subalgebra of semisimple Lie algebra, which has dimension equal to the rank, a Cartan subalgebra?

Question

Consider a real semisimple Lie algebra $\mathfrak{g}$ and a toral subalgebra $\mathfrak{h}$, which in this case is equivalent to being abelian and made of ad-semisimple elements (i.e. they are ad-diagonalizable over the complex field).

If the dimension of $\mathfrak{h}$ is equal to the rank of $\mathfrak{g}$, is it a Cartan subalgebra? (i.e. a maximal toral subalgebra with respect to inclusion)

In general, an abelian subalgebra may have dimension equal or larger than that of a Cartan subalgebra (or rank). A common example I see is that of the strictly upper-block-triangular matrices in $\mathfrak{sl}(2n,\mathbb{R})$ (also its complexification). Therefore, it is clear that in general an abelian subalgebra of a semisimple Lie algebra with dimension equal to the rank need not be a Cartan subalgebra.

However, if we add the additional requirement of semisimple elements, i.e. a toral subalgebra, not just abelian, is this still true?

Namely, is a Cartan subalgebra in a real semisimple Lie algebra a toral subalgebra of maximal dimension? If not, could you give a counterexample?

In a finite-dimensional Lie algebra over any field of characteristic $0$, the Cartan subalgebras (=nilpotent self-normalizing subalgebras) are exactly the maximal toral subalgebras. Cf. https://math.stackexchange.com/a/2497093/96384 — Torsten Schoeneberg, Jul 27 '24 at 21:09
@TorstenSchoeneberg In a finite-dimensional semisimple Lie algebra* — Callum, Jul 28 '24 at 16:02
@Callum Of course. I thought it but forgot to write it, thank you. — Torsten Schoeneberg, Jul 28 '24 at 16:39

Callum · Accepted Answer · 2024-07-26T22:13:56.223

I'm assuming by rank you mean the rank of its complexification.

In the sense that they are nilpotent subalgebras which are self-normalising, yes. In the sense that they are maximal toral, yes (if there was a larger toral subalgebra its complexification would be toral).

There is a reason we don't tend to talk about Cartan subalgebras over the reals though. They are not all conjugate unlike the complex version so our choice of one matters. Instead you talk about maximal split toral subalgebras where we require that the elements are diagonalisable over the reals. These are all conjugate but they are not in general as large as a full maximal toral subalgebra (except in the case of a split Lie algebra such as $\mathfrak{sl}(n,\mathbb{R})$). You can then extend this to a full toral subalgebra if you want.

With an appropriate Cartan involution you can split general Cartan subalgebras into a split part and a non-split part.

Note: for a compact semisimple Lie algebra there are no diagonalisable elements and we just have fully non-split toral subalgebras which are all conjugate just like in the complex case. I occasionally see them called Cartan subalgebras in this case but maximal toral subalgebra is perhaps more common still.

Is a toral subalgebra of semisimple Lie algebra, which has dimension equal to the rank, a Cartan subalgebra?

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