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](https://link.springer.com/content/pdf/10.1007/BF02863632.pdf) there is a detailed characterization and in section 5.1 of [2](https://www.sciencedirect.com/science/article/pii/S0926224501000444) there is an easy construction.



Maximal abelian subalgebras of Lie algebras over $\mathbb{C}$
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?](https://math.stackexchange.com/q/2756715/96384))


