# Hodge star isomorphism

In Petersen's Riemannian geometry text, he defines the Hodge operator $*: \Omega^k(M) \to \Omega^{n-k} (M)$ in the standard way.

He then proves (Lemma 26, Chap 7) that $*^2: \Omega^k(M) \to \Omega^k(M)$ is multiplication by $(-1)^{k(n-k)}.$

So far no problems.

However, he seems to argue that this lemma implies that the Hodge star gives an isomorphism $H^k(M) \to H^{n-k}(M),$ where we are considering the de Rham cohomology groups.

It is clear by the lemma that we have an isomorphism from $\Omega^k \to \Omega^{n-k}$ given by the Hodge star. But why must this descend to an isomorphism on cohomology?

I guess one would need to show that the Hodge star maps closed forms to closed forms, and exact forms to exact forms? Is this clear from the lemma, or am I to conclude that Pedersen is foreshadowing a theorem to come?

• Isn't this just because of functoriality? Nov 13 '14 at 21:36
• Could you expand a little on your comment? What functoriality are you referring to? Nov 13 '14 at 21:41
• Sorry, I think I was too quick. Only if one can show that the hodge stars form a chain map from $\Omega^*(M) \to \Omega^{n-*}(M)$, it would follow from functoriality, I guess. Nov 13 '14 at 21:46
• Actually, you first need to check that $\star$ maps closed/exact forms to closed/exact forms so that $\star$ induices a linear map between $H^n(M)$ and $H^{n-k}(M)$. Then , thanks to lemma 26, $\star$ has to be invertible on the cohomology level and thus an isomorphism between $H^k(M)$ and $H^{n-k}(M)$. Nov 13 '14 at 21:49
• @Bebop I agree! But is this supposed to be obvious? It's not to me, from the definition I'm given of the * Nov 13 '14 at 21:52

The Hodge star definitely does not map (for example) closed forms to closed forms. Here is a simple example to show this: on a Riemannian $n$-manifold, consider the $n$-form $\omega = f \, \text{vol}$, where $f$ is a smooth function and $\text{vol}$ is the Riemannian volume form. Note that since $\omega$ is an $n$-form, it is always closed. But $\star \omega = f$, which is closed if and only if $f$ is (locally) constant.
I don't see a good reason to expect $\star$ to give an isomorphism on cohomology, other than the appeal to the Hodge Theorem in Ted's answer. (Of course, that crucially relies on the orientability and compactness of $M$.)
Edit: It occurs to me that the point is that $\star$ interchanges $d$-closed and $d^\ast$-closed forms, i.e., $d \omega = 0$ iff $\star d \star \star \omega = \pm d^\ast (\star \omega) = 0$. Harmonic forms are precisely the forms that are both $d$-closed and $d^\ast$-closed; thus $\star$ is an automorphism of the space of harmonic forms.
No, what is obvious is that $\star$ gives an isomorphism from the space of harmonic $k$-forms ($\mathscr H^k(M)$) to the space of harmonic $(n-k)$-forms. The Hodge Theorem gives an isomorphism $H^k(M) \overset{\cong}{\to} \mathscr H^k(M)$. (I don't have Petersen's book here with me, so I can't check to see his context.)