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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 $\mathfrak{h}' = \mathfrak{a}' \oplus \operatorname{Cent}_{\mathfrak{k}}(\mathfrak{a}')$?

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    $\begingroup$ 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. $\endgroup$
    – Callum
    Nov 7, 2023 at 18:15
  • $\begingroup$ 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? $\endgroup$ Nov 8, 2023 at 10:37
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2023 at 10:40
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    $\begingroup$ 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). $\endgroup$
    – Callum
    Nov 8, 2023 at 11:50
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    $\begingroup$ 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 $\mathfrak{a}$. $\endgroup$
    – Callum
    Nov 8, 2023 at 12:05

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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.

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