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?

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.

• Thanks very much +1. I appreciate this -- also thanks for pointing out any ref. Dec 6 '21 at 21:04
• 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.
– mme
Dec 7 '21 at 14:22
• @mme Thank you! Dec 7 '21 at 14:24

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.