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Let $(M,g)$ be an oriented Riemannian manifold. Then, the codifferential $\delta$ is given by $\delta \omega=-\star d \star \omega$, where $\star$ stands for the Hodge star, $d$ for the exterior derivative and $\omega \in \Omega^1_c(M)$, where $\Omega^1_c(M)$ denotes the differential 1-forms with compact support. I am reading the prove of the following equality $$\int_M \delta(\omega) \mathrm{dvol}_g=0,$$ where $\mathrm{dvol}_g$ denotes the Riemannian volume form.

At some point it is used that: $$\int_{M} \star d \star \omega= \int_{M} d(\star \omega).$$ Why is this true?

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The left-hand side of the equality is missing $\operatorname{dvol}_g$.

For forms $\alpha, \beta$ of the same degree, we have $ (\star \alpha) \wedge \beta = (\star \beta) \wedge \alpha$ (and both are equal to $\pm \langle \alpha, \beta \rangle \operatorname{dvol}_g$, but the sign doesn't matter here).

Recall also that $\star \operatorname{dvol}_g = 1$.

Thus $(\star d \star \omega) \wedge \operatorname{dvol}_g = (\star \operatorname{dvol}_g ) \wedge d \star \omega = 1 \wedge d \star \omega = d \star \omega$. So the equality is actually pointwise.

By the way, if we now integrate over $M$, since by Stokes' theorem, $\int_M d\star \omega = 0$ (since $\omega$ is compactly supported, there is no boundary term), we have $\int_M (\star d \star \omega) \wedge \operatorname{dvol}_g = 0$. I've seen this called the divergence theorem, and I think of it as a reformulation of Stokes' theorem in terms of $\delta$.

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