Q:

The de Rham complex of a quaternion-Kahler manifold

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.

A:

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

