# Do all maximally noncompact Cartan subalgebras in a real semisimple Lie algebra arise from split toral subalgebras?

Let $$\mathfrak{g}$$ be a real, semisimple Lie algebra and $$\mathfrak{g}_{\mathbb{C}} = \mathfrak{g} + i\mathfrak{g}$$ its complexification. Following Knapp, a Cartan subalgebra of $$\mathfrak{g}_{\mathbb{C}}$$ is an abelian subalgebra $$\mathfrak{h} \subseteq \mathfrak{g}_{\mathbb{C}}$$ such that $$\operatorname{Nor}_{\mathfrak{g}_{\mathbb{C}}}(\mathfrak{h}) = \mathfrak{h}$$. A real Cartan subalgbra of $$\mathfrak{g}$$ is a subalgebra $$\mathfrak{h} \subseteq \mathfrak{g}_0$$ whose complexification $$\mathfrak{h}_{\mathbb{C}} \subseteq \mathfrak{g}_{\mathbb{C}}$$ is a Cartan subalgebra.

An example of a real Cartan subalgebra is given by starting with a Cartan involution $$\theta$$, the associated Cartan decomposition $$\mathfrak{g}= \mathfrak{k} \oplus \mathfrak{p}$$ and a maximal abelian subalgebra $$\mathfrak{a} \subseteq \mathfrak{p}$$. Then $$\mathfrak{h} := \mathfrak{a} \oplus \operatorname{Cent}_{\mathfrak{k}}(\mathfrak{a})$$ is a real Cartan subalgebra. In fact $$\mathfrak{h}$$ is $$\theta$$-invariant and maximally noncompact. In particular, $$\mathfrak{a}$$ is $$\mathbb{R}$$-split, meaning $$\operatorname{ad}_{\mathfrak{g}}(\mathfrak{a})$$ is diagonalizable.

Question: If I start with an abelian $$\mathbb{R}$$-split subalgebra $$\mathfrak{a}' \subseteq \mathfrak{g}$$ that is maximal among the abelian $$\mathbb{R}$$-split subalgebras (one may call $$\mathfrak{a}'$$ a maximal split toral subalgebra). Can I then find a Cartan subalgebra $$\mathfrak{h}'$$ that contains $$\mathfrak{a}'$$? Moreover, after a conjugation, can I suppose that $$\mathfrak{h}'$$ is $$\theta$$-invariant and maximally non-compact and $$\mathfrak{h}' = \mathfrak{a}' \oplus \operatorname{Cent}_{\mathfrak{k}}(\mathfrak{a}')$$?

• Cartan subalgebra here coincides with maximal toral subalgebra as it does in the complex case so any toral subalgebra can be extended to a CSA. You can definitely also assume that your $\mathfrak{h}'$ is $\theta$-stable up to conjugation. Indeed I think all "maximally noncompact" aka "maximally split" CSA's are conjugate. If this isn't explained in Knapp it will be somewhere in Satake's classification of real simple Lie algebras. Nov 7, 2023 at 18:15
• In Knapp it is explained that every CSA is conjugated to a $\theta$-stable one. And that all the maximally noncompact CSA are conjugate. In the complex case, Knapp shows that CSA = maximal abelian diagonalizable = maximal toral. But how do I see this in the real case? Nov 8, 2023 at 10:37
• Maximally noncompact CSA is defined as in the post ($\mathfrak{a}$ being maximal abelian subalgebra of $\mathfrak{p}$). Maximally split CSA is defined as a CSA that contains a split toral subalgebra $\mathfrak{a}'$. How do I prove that these two notions coincide? For a starters, why can it not be that $\operatorname{dim}(\mathfrak{a}') > \operatorname{dim}(\mathfrak{a})$? If $\mathfrak{a}'$ would have to lie in $\mathfrak{p}$, then yes, but since it does not have to, I don't know how to show it. Nov 8, 2023 at 10:40
• For the first part, it is toral because it is contained in its complexification which is toral. If it isn't maximal there is some $X$ which commutes with every element. But then $X$ must commute with all elements of the complexification as well so we have a contradiction. Of course in the real case these are not equivalent to maximal abelian diagonalisable (that being equivalent to maximal split toral instead). Nov 8, 2023 at 11:50
• For the second part, I see now that was your main question. If we already know that all our CSA's are conjugate to $\theta$-stable ones then after conjugating to $\theta$-stable $\mathfrak{h}'$, say, we know it splits into parts in $\mathfrak{k}$ and $\mathfrak{p}$: $\mathfrak{h}' = \mathfrak{h}'_{\mathfrak{k}} \oplus \mathfrak{h}'_{\mathfrak{p}}$. Conjugation will not change diagonalisabilty so the split part $\mathfrak{a}'$ must have become $\mathfrak{h}'_{\mathfrak{p}}$ which is abelian so has dimension less than that of $\mathfrak{a}$. Nov 8, 2023 at 12:05

Thank you for clearing things up @Callum.

Let me start with a definition. Let $$\mathbb{K}$$ be a field and $$\mathfrak{g}$$ be a finite dimensional $$\mathbb{K}$$-Lie algebra. A subalgebra $$\mathfrak{h}\subseteq \mathfrak{g}$$ is called toral, if $$\mathfrak{h}$$ is abelian and the linear maps in $$\operatorname{ad}(\mathfrak{h}) \subseteq \mathfrak{gl}(\mathfrak{g})$$ are diagonalizable over the algebraic closure of $$\mathbb{K}$$. If $$\operatorname{ad}(\mathfrak{h})$$ is moreover diagonalizable over $$\mathbb{K}$$, then $$\mathfrak{h}$$ is $$\mathbb{K}$$-split toral. If $$\mathfrak{h}$$ is maximal among the $$\mathbb{K}$$-split toral subalgebras, then it is called maximal $$\mathbb{K}$$-split toral.

Let now $$\mathfrak{g}$$ be a real, semisimple, finite-dimensional Lie algebra and $$\mathfrak{a}'$$ a maximal $$\mathbb{R}$$-split toral subalgebra. Now let $$\mathfrak{h}'$$ be a maximal toral subalgebra containing $$\mathfrak{a}'$$. This exists because of finite-dimensionality.

To prove that $$\mathfrak{h}'$$ is a real Cartan subalgebra, we have to show that $$\mathfrak{h}'_\mathbb{C}$$ is a complex Cartan subalgebra. It is easy to show that $$\mathfrak{h}'_\mathbb{C}$$ is abelian. Since $$\mathfrak{h}'$$ is abelian and the elements of $$\operatorname{ad}(\mathfrak{h}')$$ are $$\mathbb{C}$$-diagonalizable, they are in fact simultaneously diagonalizable. Hence for any $$X,X' \in \mathfrak{h}'$$, $$\operatorname{ad}(X+iX') = \operatorname{ad}(X) + i\operatorname{ad}(X') \in \mathfrak{gl}(\mathfrak{g}_\mathbb{C})$$ are diagonalizable. This means that $$\mathfrak{h}'_{\mathbb{C}}$$ is a toral subalgebra of $$\mathfrak{g}_\mathbb{C}$$. In fact, $$\operatorname{ad}(\mathfrak{h}'_{\mathbb{C}})$$ is simultaneously diagonalizable and $$\mathfrak{h}'_\mathbb{C}$$ is maximal toral. Then by [Knapp, Prop 2.13], $$\mathfrak{h}'_\mathbb{C}$$ is a Cartan subalgebra of $$\mathfrak{g}_\mathbb{C}$$. Hence $$\mathfrak{h}'$$ is a Cartan subalgebra of $$\mathfrak{g}$$, answering the first question.

Now that we have a real Cartan subalgebra $$\mathfrak{h}' \supseteq \mathfrak{a}'$$, we can apply [Knapp, Prop 6.59] to conjugate it to a $$\theta$$-stable Cartan subalgebra $$\mathfrak{h} = \operatorname{Ad}(g)(\mathfrak{h}')$$. Let $$\mathfrak{a} := \mathfrak{h}\cap \mathfrak{p}, \mathfrak{m} := \mathfrak{h}\cap \mathfrak{k}$$. Since $$\mathfrak{h}$$ is $$\theta$$-invariant, $$\theta(\mathfrak{a}) \subseteq \mathfrak{a}$$ and $$\theta(\mathfrak{m}) \subseteq \mathfrak{m}$$, so $$\mathfrak{h} = \mathfrak{a} \oplus \mathfrak{m}$$. Note that since the elements of $$\operatorname{ad}(\mathfrak{a}')$$ are $$\mathbb{R}$$-diagonalizable, so must the conjugated elements in $$\operatorname{Ad}(g)(\mathfrak{a}') \subseteq \mathfrak{h}$$ be. We know that for $$X\in \mathfrak{a} \subseteq \mathfrak{p}, Y \in \mathfrak{m} \subseteq \mathfrak{k}$$, $$\operatorname{ad}(X)$$ is $$\mathbb{R}$$-diagonalizable, but $$\operatorname{ad}(Y)$$ is not. In fact $$\operatorname{ad}(X+Y)$$ is $$\mathbb{R}$$-diagonalizable if and only if $$Y = 0$$. Therefore $$\operatorname{Ad}(g)(\mathfrak{a}') \subseteq \mathfrak{a}$$. $$\mathfrak{a}$$ is split toral, and since $$\operatorname{Ad}(\mathfrak{a}')$$ is maximal split toral, we have $$\operatorname{Ad}(g)(\mathfrak{a}') = \mathfrak{a}$$.

Since in any $$\theta$$-stable CSA $$\mathfrak{h}'' = \mathfrak{a}'' \oplus \mathfrak{m}''$$, $$\mathfrak{a}''$$ is a split toral subalgebra, $$\mathfrak{h}''$$ is maximally noncompact (i.e. $$\operatorname{dim}(\mathfrak{a}'')$$ is maximal) if and only if it contains a maximal $$\mathbb{R}$$-split toral subalgebra.

• I don't have Knapp's Prop. 2.13 in front of me right now, so I don't know how much his proof uses the base field $\mathbb C$, but in fact it is true for any semisimple Lie algebra over any field of characteristic $0$ that the maximal toral subalgebras are exactly the Cartan subalgebras. That's exercise 3 to vol VII, §2 of Bourbaki's Lie groups and Lie algebras. Cf. math.stackexchange.com/q/3480970/96384, math.stackexchange.com/q/3537312/96384, as well as the many links to and from math.stackexchange.com/q/1071417/96384. Nov 9, 2023 at 20:28