 Solution

On an oriented Riemannian $n$-manifold, the metric volume form is the unique top form taking value one on every positively oriented orthonormal tangent basis. In a positive coordinate chart it isFor another positive coordinate chart $y$, write $A=\partial x/\partial y$. Then $g_y=A^{\mathsf{T}}{g}_{x}A$, so $\sqrt{\det g_y}=\det A\sqrt{\det g_x}$, since $\det A>0$. The wedge of coordinate differentials acquires the same determinant factor. The two local expressions agree, proving coordinate independence and smooth global existence. Positivity of the chosen charts is what fixes the sign of the volume form.

The metric induces the inner product on exterior powers of a cotangent space. Define the Hodge star operator $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ byfor forms of the same degree, and apply it fiberwise to smooth forms. In an oriented orthonormal coframe, if $I$ is an increasing $p$-index and $I^c$ its ordered complement, $\ast e^I=ϵ(I,I^c)e^{I^c}$. Exchanging the two blocks has sign $(-1{)}^{p(n-p)}$, so

Use the real $L^2$ inner product $(\alpha ,\beta )={\int }_M\alpha \wedge \ast \beta$. For a $p$-form $\alpha$ and a $(p+1)$-form $\beta$, the graded Leibniz rule and the given boundaryless Stokes identity giveThe codifferential acting on $\beta$ has sign $(-1{)}^{n(p+2)+1}$. Applying the star once more and using its square on the $(n-p)$-form $d\ast \beta$ givesThe exponent reduction uses parity, since $p^2\equiv p(\mathrm{mod}2)$. Combining the two displayed identities proves $(d\alpha ,\beta )=(\alpha ,\delta \beta )$. Symmetry of the real inner product supplies preciselyThus Hodge integration by parts in arbitrary degree identifies $\delta$ as the formal adjoint of $d$. The displayed bilinear identity extends to complex differential forms by complex linearity. For positive norm identities on complex forms, use the Hermitian inner product with conjugation instead.

With the nonnegative convention, the Hodge Laplacian isIt is the form-valued extension of the Laplace-Beltrami operator, whose action on functions is $-\mathrm{div}\mathrm{grad}$. A harmonic differential form is a smooth form in $\ker \Delta$. The adjoint identity givesIf $\Delta \alpha =0$, both norms vanish, and smoothness impliesThe closed compact hypotheses are essential for the absence of boundary terms and this global energy conclusion.

The Hodge decomposition theorem states that on a closed oriented Riemannian manifoldis an $L^2$-orthogonal direct sum, with finite-dimensional harmonic space ${\mathcal{H}}^p$. Every smooth form has unique harmonic, exact and coexact components, although primitives of the latter two need not be unique. Equivalently every de Rham cohomology class has a unique harmonic representative. An exact harmonic form belongs to both the harmonic and exact summands; their orthogonality makes its squared norm zero. Therefore $\boxed{\alpha \text{ exact and harmonic}⟹\alpha =0}$. The same conclusion follows directly from $\alpha =d\eta$ and $\Vert \alpha {\Vert }^2=(\delta \alpha ,\eta )=0$.