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As we all know, for a complex manifold $M$, its de Rham complex admits a decomposition into a double complex called the Dolbeault complex. If $M$ also admits a Kahler metric, then we get the wonderful Kahler identities.

Can I ask what extra structure we get on $\Omega(M)$ if we also assume the existence of a quaternion-Kähler structure? I assume here that $M$ has non-zero Ricci curvature, since in the flat case it is well-known that we get a representation of the quaternions of $\Omega(M)$, which is to say, we get a hyper-Kahler manifold.

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    $\begingroup$ I think this is discussed in Simon Salamon's book "Riemannian geometry and holonomy groups". The point is that, just as in the other cases of special holonomy, the Laplacian commutes with the qK type decomposition (this is actually an old theorem due to S.S. Chern), and you get a number of identities that are analogs of the Kahler identities. See the book for details. $\endgroup$
    – Robert Bryant
    Commented Jan 31, 2014 at 15:36

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The fundamental 4-form on a quaternionic-Kahler manifold is closed and gives the same kind of decomposition as the Kahler form on a Kahler manifold. Reference: Bonan, Edmond, Sur l'algèbre extérieure d'une variété presque hermitienne quaternionique. [Exterior algebra of a quaternionic almost Hermitian manifold] C. R. Acad. Sci. Paris Sér. I Math. 295 (1982), no. 2, 115–118; Semmelmann, Uwe; Weingart, Gregor, Vanishing theorems for quaternionic Kähler manifolds, J. Reine Angew. Math. 544 (2002), 111–132, http://arxiv.org/abs/math/0001061

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