 Is the wedge product of two harmonic forms on a compact Riemannian manifold harmonic? I'm looking for a counterexample that the textbooks say exists.
 I would like to see a counter example that is on a complex manifold, Ricciflat (or Einstein) manifold or both, if it is at all possible.
 In general, I'm trying to understand the interaction between the wedge product, Hodge star and the Laplacian on forms and it's eigenvectors, references will be much appreciated.



It is easy to construct examples on Riemann surfaces of genus >1. Take any surface like this. Let A and B be two harmonic 1forms, that are not proportional. Then A \wegde B is nonzero, but it vanishes at some point, since both A and B have zeros. At the same time a harmonic 2 form on a Rieamann surface is constant. Explicite examples of 1forms on Rieamann surfaces can be obtained as real parts of holomorphic 1forms. Note of course that the above example is complex, and Einstein just take the standard metric of curvature 1. If you want an example on a Ricci flat manifold you should take a K3 surface. It is complex and admits a Ricci flat metric. Now, its second cohomology has dimesnion 22. Now it should be possible to find two antiselfdual twoforms whose wedge product vanishes at one point on K3 but is not identically zero. This is because the dimesnion of the space of self dual forms is 19 which is big enought to get vanishing at one point 


Interestingly in (24) of hepth/9603176 it is mistakenly claimed that the wedge product of harmonic forms is automatically harmonic. Because it is false we still do not know the predicted existence of those middle dimensional $L^2$ harmonic forms on these noncompact complete hyperkahler manifolds. 


Generically, the wedge product of two harmonic forms will not be harmonic. It is harder to find examples than counterexamples. For example, on compact Lie groups with a biinvariant metric or, more generally, on riemannian symmetric spaces, harmonic forms are invariant and invariance is preserved by the wedge product. In general, though, this is not the case. According to Kotschick (see, e.g., this paper) manifolds admitting a metric with this property are called geometrically formal and their topology is strongly constrained. He has examples, already in dimension 4, of manifolds which are not geometrically formal. 


Here's a counterexample from the theory of nilmanifolds, which by their very nature are not formal. Take a compact quotient $H^3/\Gamma$ of the Heisenberg group. It admits invariant 1forms $e^1,e^2,e^3$ with $de^1=0=de^2$ and $de^3=e^1\wedge e^2$. Then $e^1,e^2$ are harmonic, but $e^1\wedge e^2$ is exact, so not harmonic. You can take a product with $S^1$ to get a complex (nonKähler) surface on which the same thing works, but not I am afraid Ricciflat or Einstein. 


Using homological perturbation theory, one can repair this defect. More precisely, on the space of harmonic forms, there is an $A_\infty$ structure with no differential whose 2ary operation (multiplication) is constructed by wedging two harmonic forms then projecting the result back to the space of harmonic forms. See "Strong homotopy algebras of Kahler manifolds" by S.A. Merkulov (Int. Math. Res. Lett. no. 3 153164) for details of the construction. EDIT: Also, if the manifold is compact then the natural inclusion of harmonic forms into arbitrary forms becomes an equivalence of $A_\infty$ algebras, where the space of all forms has its usual dgalgebra structure. 

