Hodge map and the Cohomology Ring of a Riemannian Manifold

Question

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.)

Answer 1

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.

Answer 2

As has been observed in the comments, it suffices to construct a closed differential form $\omega$ for which $\star\omega$ is not closed. Here is an explicit example.

Let $M$ be the circle $S^1$ for which we take the convenient model $\mathbf R/\mathbf Z$. Let $g$ be any $1$-periodic strictly positive nonconstant smooth function on $\mathbf R$, for example $g(x)=\cos(2\pi x)^2+1$. Then, $g\,dx\otimes dx$ is a Riemannian metric on $M$. The associated volume form on $M$ is $\mathrm{vol}=\sqrt g \,dx$.

Let $\omega$ be the $1$-form $dx$ on $M$. Of course, $\omega$ is closed. The Hodge dual $\star\omega$ is the smooth function determined by $$ \omega\wedge\star\omega=\langle\omega,\omega\rangle\mathrm{vol}, $$ i.e., $$ \star\omega \cdot dx= \frac1g \cdot\sqrt g \,dx=\frac1{\sqrt g}dx. $$ It follows that $\star\omega$ is the function $1/\sqrt g$, which is not closed since $g$ is not constant.