Hypercomplex structure as integrable $Gl(\mathbb{H},n)$-structure. Although there is some mess with notation, we say that a manifold $M^{4n}$ has an almost hypercomplex structure if there are three almost complex structures $I,J,K$ satisfing quaternions relations. Now classical definition of hypercomplexity requires all of them to be integrable. Since an almost hypercomplex structure is nothing but $Gl(\mathbb{H},n)$ structure, I would like to know if being hypercomplex is equivalento to be an integrable (in the sence of G-structur) almost hypercomplex? Ofcourse being integrable implies integrability of $I,J,K$ yet I can't see if the seperate integrability of them implies that they can be simultaniously presented in a canonical way in some chart?
 A: The answer is yes due to the existence of the Obata connection:
Theorem [Obata 1955]. Let $(M, I, J, K)$ be a hypercomplex manifold, (i.e.: $I$, $J$ and $K$ are integrable almost complex structures satisfying the quaternionic relations). Then there exists a unique torsion-free connection $\nabla$ such that $\nabla I = \nabla J = \nabla K = 0$.
Do you agree?
Here are some more explanations: In general, the integrability of a $G$-structure is equivalent to the existence of an compatible torsion-free connection. There are several ways to express compatibility of a connection with a $G$-structure, one of them being that parallel transport with respect to the connection respects the $G$-structure. With this definition it is clear that the Obata connection is compatible with the $GL(n,\mathbb{H})$-structure defined by $I$, $J$ and $K$. If you want to learn more about integrability of $G$-structures and the relation with compatible connections, I suggest you take a look at these notes, especially section 4.2: Torsion and curvature (and their relevance to integrability).
