If $M=G/H$ is a reductive homogeneous space then we can write $\frak{g}=\frak{m}+\frak{h}$
where $[\frak{h}, \frak{m}]\subset \frak{m}$. Here $\frak{g}$ and $\frak{h}$ are the Lie algebras of $G$ and $H$ respectively. Let $T\_o$ be the torsion of the canonical connection (of the second kind) w.r.t. this decomposition at $o=\pi(e)$. 


My questions is: when can we decompose $\frak{m}$ as 
$$\frak{m}=\frak{m}\_0+\frak{m}\_1$$
where $T\_o(X\_1,X\_2)=0$ for $X\_1,X\_2\in \frak{m}\_0$, $T\_o(Y\_1,Y\_2)\ne 0$ for $Y\_1,Y\_2\in \frak{m}\_1$, and $T\_o(X,Y)=0$ for $X\in \frak{m}\_0$ and $Y\in \frak{m}\_1$? For example, any compact connected Lie group $G$ is a semidirect product of its commutator subgroup $G^\prime$ and an Abelian subgroup $G\_0$. We can therefore take $\frak{m}\_0$ to be the Lie algebra of $G\_0$ and $\frak{m}\_1=[\frak{g},\frak{g}]$, the Lie algebra of $G^\prime$. It's well known that the torsion for $G$ is given by $T\_e(X,Y)=-[X,Y]$. 



Decomposition of Lie subspaces
I am starting to think that you can always do this, but perhaps I am missing something at this later hour.


The torsion of the canonical connection at the origin defines a skewsymmetric bilinear map $T\_0: \Lambda^2 \mathfrak{m} \to \mathfrak{m}$ by
$$
T\_0(X,Y) = -[X,Y]\_{\mathfrak{m}}
$$
where the subscript indicates the natural projection $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m} \to \mathfrak{m}$. In turn this defines a linear map $\tau : \mathfrak{m} \to \operatorname{End}(\mathfrak{m})$ by
$$
\tau(X) = - [X, -]\_{\mathfrak{m}}
$$


We define $\mathfrak{m}\_0 = \ker \tau$. I claim that $\mathfrak{m}\_0$ is an $\operatorname{Ad}(H)$-module. I assume $H$ is connected, so it is enough to show that if $\tau(X) = 0$, then for all $Z \in \mathfrak{h}$, $\tau([Z,X])=0$. This follows from the following calculation (using the Jacobi identity of $\mathfrak{g}$):
$$
\tau([Z,X]) = [[Z,X],-]\_{\mathfrak{m}} = [Z,[X,-]]\_{\mathfrak{m}} - [X,[Z,-]]\_{\mathfrak{m}}
$$
The first term in the RHS vanishes because it can be rewritten as
$$[Z,[X,-]\_{\mathfrak{m}}]\_{\mathfrak{m}}$$
and we use that $\tau(X) = 0$. The second term in the RHS vanishes because of reductivity and the fact that $\tau(X)=0$, since it can be rewritten as
$$-[X,[Z,-]\_{\mathfrak{m}}]\_{\mathfrak{m}}$$


By reductivity again, $\mathfrak{m}\_0$ has an $\operatorname{Ad}(H)$-module complement $\mathfrak{m}\_1$ in $\mathfrak{m}$.


