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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?

J.E.M.S
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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).

Seub
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  • Actually I knew about existance of Obata's connection, yet generally $\nabla t$=0 for torsion free connection and $t$ almost complex implies t integrable (constant coeficients in some chart reper) It is not clear to me that if some almost complex tensors are paralale with respect to the same torsion free conection they are simulatanous trivializable in common chart. – J.E.M.S Nov 20 '16 at 12:19
  • I edited my answer to include more explanations. Let me know if they clarify things enough. – Seub Nov 21 '16 at 03:48
  • Actually "In general, the integrability of a G-structure is equivalent to the existence of an compatible torsion-free connection" is almost never true, I think there is a paper of an old japanese geometry school where it is show that 1-integrability (existance of compatybil torsion free connection) implies integrability in case of 5 or 6 groups in Gl(n). for Example compatybil torsion free connection always exists for Riemann structure yet it is not always integrable. – J.E.M.S Nov 21 '16 at 15:56
  • Mmh, I think that maybe what you are talking about applies to connections in the principal bundle associated to the G-structure, but here I am talking about connections in the tangent bundle. I could be wrong about that, but anyway what I wrote is proven in the notes that I mention, see Corollary 4.40. Another remark: I think you should take a look at this paper, maybe it will make you happy: http://www.bdim.eu/item?fmt=pdf&id=RLIN_1993_9_4_1_43_0 – Seub Nov 21 '16 at 16:10