We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology? Namely,
Is the Hodge dual of a closed form also a closed form?
Is the Hodge dual of an exact form also an exact form?
Is there a Hodge dual of de Rham cohomology and singular cohomology?