For a compact Riemannian manifold $M$, we know that the Hodge map $\ast$ and Laplacian $\Delta$ commute. From Hodge decomposition and its implied isomorphism between harmonic forms and cohomology classes we now that an indued on the cohomology ring $H(M)$, which is usually denoted again by $\ast$.


Now one could also naively define a map
$$
\ast:H(M) \to H(M), ~~~~~~~~~~~~~ [\omega] \mapsto [\ast(\omega)],
$$
and hope that they coincide. The thing is, as far I can tell, this map doesn't even seem to be well-defined. Can someone please prove whether or not it is well-defined. 


(In the case that it is not well-defined, any philosophical motivation for why the naive approach doesn't work would be appreciated.)



Hodge map and the Cohomology Ring of a Riemannian Manifold
You've probably seen this before, but in case you haven't:


Let $M$ be a compact oriented manifold. Let $\Omega^k$ be the smooth $k$-forms, let $Z^k$ be the closed $k$-forms and let $B^k$ be the exact $k$-forms. So $H^k=Z^k/B^k$. As Johannes's answer shows, $\ast$ does not carry $Z^k$ to $Z^{n-k}$, nor $B^k$ to $B^{n-k}$. (For an example of the latter, take $S^1$ with its usual metric, and $f:S^1 \to \mathbb{R}$ any non constant function. Then $df = f'(\theta) d \theta$ is exact, but $\ast(df) = f'(\theta)$ is not. ) So you cannot define a map $H^k \to H^{n-k}$ in your naive way.


We can put an inner product on $\Omega^k$ by $\langle \omega, \eta \rangle = \int\_M \omega \wedge \ast(\eta)$. If we pretend that everything works like finite dimensional vector spaces (and the hard part of Hodge theory is justifying this), then $\Omega^k$ splits into $B^k \oplus (B^\*)^k \oplus A^k$. Here $(B^\*)^k$ is $(Z^k)^{\perp}$ and $A^k = (B^k \oplus (B^\*)^k)^{\perp}$. The space $A^k$ is the harmonic forms. Notice that $Z^k = A^k \oplus B^k$, so $H^k = Z^k/B^k \cong A^k$.


Hodge star takes $A^k \to A^{n-k}$, $B^k \to (B^\*)^{n-k}$ and $(B^\*)^k \to B^{n-k}$. So $B^k$ doesn't go to $B^{n-k}$ and $Z^k = A^k \oplus B^k$ doesn't go to $Z^{n-k} = A^{n-k} \oplus B^{n-k}$, making your naive strategy not work. But we do have $Z^k/B^k \cong A^k \to A^{n-k} \cong Z^{n-k}/B^{n-k}$, giving us the Hodge map.


