Hodge dual of de Rham cohomology and singular cohomology 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?
 A: 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.
A: The Hodge dual of a closed form is a co-closed form (meaning, in the kernel of $d^{\ast}$), and the Hodge dual of an exact form is a co-exact form (meaning, in the image of $d^{\ast}$). Both these facts follow from the identity $\ast \circ d = (-1)^{k+1} d^{\ast} \circ \ast$. (I hope I got the sign right.)
As a basic example, take the two dimensional torus $\mathbb{R}^2/\mathbb{Z}^2$ with the standard Euclidean metric inherited from $\mathbb{R}^2$. Consider a $1$-form $\alpha = f(x,y) dx + g(x,y) dy$. Then $d \alpha = \left(\tfrac{\partial g}{\partial x} - \tfrac{\partial f}{\partial y} \right) dx \wedge dy$, so $\alpha$ is closed if and only if $\tfrac{\partial g}{\partial x} - \tfrac{\partial f}{\partial y}=0$. Meanwhile (I might be off by a sign), we have $\ast(\alpha) = f(x,y) dy - g(x,y) dx$, so $\ast(\alpha)$ is closed if and only if $\tfrac{\partial f}{\partial x} + \tfrac{\partial g}{\partial y}=0$. So the first of these corresponds to "curl is 0" and the other corresponds to "div is 0".
However, if our $n$-manifold $M$ is compact (and oriented, but we need that to define $\ast$ in the first place), we have the Hodge theorem, which tells us that each cohomology class has a unique harmonic representative, which is both closed and co-closed. So Hodge star is an isomorphism from the harmonic representatives of $H_{DR}^k(M)$ to the the harmonic representatives of $H_{DR}^{n-k}(M)$.
This isomorphism depends on the metric on $M$, so it doesn't have a purely topological description. It does have a topological consequence though, namely, a proof of Poincare duality. Poincare duality says that $\langle \alpha, \beta \rangle = \int_M \alpha \wedge \beta$ is a perfect pairing between $H_{DR}^k(M)$ and $H_{DR}^{n-k}(M)$. To this end, it is enough to show that, for each nonzero harmonic $k$-form $\alpha$, there is a harmonic $(n-k)$-form $\beta$ with $\int_M \alpha \wedge \beta\neq 0$. Indeed, it turns out that $\int_M \alpha \wedge \ast(\alpha) > 0$ for any nonzero $\alpha$. If I recall correctly, Voisin's Hodge theory book proves Poincare duality this way as a warm up, before proving Serre duality by a harder version of this argument.
