Consider a product manifold $X=P \times P_{\perp}$, with $P$ and $P_{\perp}$ cycles of $X$, and let $\omega$ be the associated differential form of $P$ in the sense of de Rham duality (I suppose the correct words here are something like "let $\omega$ be a harmonic representative of the cohomology class of $P$").
Let $P^*$ be a cycle in the cohomology class which is de Rham dual to the Hodge dual of $\omega$, i.e to $* \omega$.
Is it generally true $P^*$ and $P_{\perp}$ are in the same cohomology class?
What I actually care about knowing is if $P^*$ and $P_{\perp}$ necessarily have the same volume.
The answer seems to be yes in the trivial example of a torus $X=T^2$, because the cohomology class of 1-cycles for which the 1-form $\omega=dx$ is a reprentatitive, is perpendicular to the cycles which are in the de Rham dual of $*\omega=dy$.
Further can any analogous statement can be made for manifolds which are not product manifolds?
I apologize for the very basic question (and any imprecisions in its formulation) - I have very little knowledge of cohomology. An answer which adressed my imprecisions in langugage would be especially instructive.