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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?

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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.

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    $\begingroup$ Thanks very much +1. I appreciate this -- also thanks for pointing out any ref. $\endgroup$
    – wonderich
    Dec 6 '21 at 21:04
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    $\begingroup$ For these signs I always check Warner. The adjoint is defined on p-forms on an n-manifolds as $d^* = (-1)^{n(p+1)+1} *d*$ and one has $** = (-1)^{p(n-p)}$, so $$d^* * = (-1)^{n(n-p+1)+1} (-1)^{p(n-p)} *d = (-1)^{p+1} * d,$$ I think. $\endgroup$
    – mme
    Dec 7 '21 at 14:22
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    $\begingroup$ @mme Thank you! $\endgroup$ Dec 7 '21 at 14:24
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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.

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