Is the wedge product of two harmonic forms harmonic? 
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*Is the wedge product of two harmonic forms on a compact Riemannian manifold harmonic? I'm looking for a counter-example that the textbooks say exists. 

*I would like to see a counter example that is on a complex manifold, Ricci-flat  (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 eigen-vectors, references will be much appreciated.

 A: Interestingly in (24) of hep-th/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 non-compact complete hyperkahler manifolds. 
A: Here is a cop-out answer (sorry in advance). Consider the 0-form $f(x)=x$ on $\mathbb{R}$. Clearly $f$ is harmonic, but $(f\wedge f)_{x}=x^2$ isn't. Since the wedge product is just the point-wise product on 0-forms, this is a valid example!
A: 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 1-forms $e^1,e^2,e^3$ 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 (non-Kähler) surface on which the same thing works, but not I am afraid Ricci-flat or Einstein. 
A: It is easy to construct examples on Riemann surfaces of genus $>1$. Take any surface like this. Let $A$ and $B$ be two harmonic $1$-forms, that are not proportional. Then $A \wedge B$ is non-zero, but it vanishes at some point, since both $A$ and $B$ have zeros. At the same time a harmonic $2$-form on a Riemann surface is constant.
Explicit examples of $1$-forms on Riemann surfaces can be obtained as real parts of holomorphic $1$-forms.
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 dimension $22$. Now it should be possible to find two anti-self-dual two-forms  whose wedge product vanishes at one point on $K3$ but is not identically zero. This is because the dimension of the space of self dual forms is 19 which is big enought to get vanishing at one point  
A: 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 2-ary 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. 3 153--164) 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 dg-algebra structure.
A: Generically, the wedge product of two harmonic forms will not be harmonic.  It is harder to find examples than counter-examples.  For example, on compact Lie groups with a bi-invariant 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.
