Maximal abelian subalgebras of Lie algebras over $\mathbb{C}$

Question

Let $\mathfrak g$ be the Lie algebra of a compact connected Lie group $G$. Let $\mathfrak g_{\mathbb{C}}$ be the complexification of $\mathfrak g$ and let $\mathfrak h \subset \mathfrak g_{\mathbb{C}}$ be a complex Lie subalgebra satisfying $\mathfrak h + \overline{\mathfrak h} = \mathfrak g_{\mathbb{C}}$. Suppose that $\mathfrak a \subset \mathfrak h$ is a maximal abelian Lie subalgebra of $\mathfrak h$. Does it hold that $\mathfrak a + \overline{\mathfrak a}$ is a maximal abelian subalgebra of $\mathfrak g_{\mathcal C}$?

Edit: A nontrivial example:

Suppose that $G$ is an even dimensional compact Lie group and suppose that is endowed with a left-invariant complex structure(*). Take $\mathfrak h$ as the set of all left-invariant vector fields that annihilates every local holomorphic function on $G$.

(*): This kind of complex structure always exist. In Proposition 2.5 of 1 there is a detailed characterization and in section 5.1 of 2 there is an easy construction.

Answer 1

As discussed in the comments, the answer is no, for a striking reason: In general, $\mathfrak{a} + \overline{\mathfrak{a}}$ is not even a subalgebra.

As example, take $\mathfrak{g} = \mathfrak{su}_2$ which we can e.g. (following Wikipedia) define as the three-dimensional real Lie algebra with basis $u_1, u_2, u_3$ satisfying

$$\left[u_3, u_1\right] = 2u_2, \quad \left[u_1, u_2\right] = 2u_3, \quad \left[u_2, u_3\right] = 2u_1.$$

It is well-known that the complexification $\Bbb C\otimes\mathfrak{g}$ is isomorphic (as complex Lie algebra) to $\mathfrak{sl}_2(\Bbb C)$. Concretely, an isomorphism is given by $$f(-\frac{i}{2}\otimes u_1 -\frac{1}{2}\otimes u_2) = \pmatrix{0&1\\0&0},$$ $$ f(-\frac{i}{2}\otimes u_1 +\frac{1}{2}\otimes u_2)=\pmatrix{0&0\\1&0},$$ $$ f(-i\otimes u_3) = \pmatrix{1&0\\0&-1}$$

One sees from this that $f$ identifies the conjugation action $a\otimes g \mapsto \bar a\otimes g$ on $\Bbb C\otimes \mathfrak{g}$ with $$\pmatrix{a&b\\c&-a} \mapsto \pmatrix{-\bar a&-\bar c\\-\bar b&\bar a}$$ on $\mathfrak{sl}_2(\Bbb C)$ (and $\mathfrak{g}$ with the fixed point set of that, the traceless skew-hermitian matrices, in particular $u_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \quad u_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad u_3 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$ ).

But then if we choose $\mathfrak{h} = \mathfrak{g}_{\Bbb C}$, and $\mathfrak{a}$ as the one-dimensional algebra generated by $\pmatrix{0&1\\0&0}$ (which is maximal abelian in $\mathfrak{h}$), we have $\overline{\mathfrak{a}} = \pmatrix{0&0\\*&0}$, hence $$\mathfrak{a} + \overline{\mathfrak{a}} = \lbrace \pmatrix{0&b\\c&0}: b,c\in \Bbb C\rbrace$$ which is not closed under the Lie bracket, hence no Lie subalgebra of $\mathfrak{g}_{\Bbb C}$.

NB. If (in the special case $\mathfrak{h} = \mathfrak{g}_{\Bbb C}$) $\mathfrak{a}$ is chosen as a Cartan subalgebra of $\mathfrak{g}_{\Bbb C}$, in some low-dimensional examples it seems to me that $\overline{\mathfrak{a}} = \mathfrak{a}$ and hence the assertion is true. Maybe one can go from there to a case of more general $\mathfrak{h}$. However, I am not even sure if/why this should hold in the special case. (I've asked this as a new question MSE/2756715: Are all CSA's in $\mathfrak{g}_{\Bbb C}$ invariant under the conjugation action?)

Answer 2

I am sorry to tell I believe the answer is No.

Consider $\mathfrak g = \mathfrak{su}(2)$ and its complexification $\mathfrak g_{\mathbb C} = \mathfrak{sl}(2,\mathbb C)$, and $\mathfrak h= \mathfrak{sl}(2,\mathbb R)$. Now look at $\mathfrak a=\mathfrak{so}(2)$, the set of antisymmetric matrices in $\mathfrak g$. Then $\mathfrak a+\bar{\mathfrak a}$ is not maximal Abelian in $\mathfrak g_\mathbb C$, because it is properly contained in $\mathfrak a +\bar{\mathfrak a} + \mathbb C\begin{pmatrix} i \\ 1 \\ -1 \\ -i \end{pmatrix}$. There are more example like this.