TITLE: Hodge dual of de Rham cohomology and singular cohomology
QUESTION [8 upvotes]: 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?

REPLY [8 votes]: The Hodge * operator action on cohomology is generally speaking
metric-dependent, hence * is not well-defined without fixing the metric.
There are some caveats. On complex curves, for example, the
Hodge * operator is complex rotation, which depends only on
complex structure. This gives an example of a manifold
for which the action of * is metric-dependent: indeed, the
action of complex rotation on $H^1$ determines the
biholomorphism class of the complex  curve (Torelli).
On compact 2n-dimensional manifolds,
the *-operator in the middle cohomology is determined by
the conformal structure.