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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 $\Vert \alpha {\Vert }^2=(\delta \alpha ,\eta )=0$.

Use the given orientation throughout, and the usual convention that the manifold has no boundary. In an oriented manifold chart the metric volume form isIf $y$ is another oriented chart and $A=\partial x/\partial y$, then $g_y=A^t{g}_{x}A$, so $\sqrt{\det g_y}=(\det A)\sqrt{\det g_x}$ because $\det A>0$. This is exactly the transformation of the top exterior product. The formulas therefore agree on overlaps and define a smooth positive volume form. Equivalently it takes value one on any positively oriented orthonormal tangent frame.

The metric induces an inner product on $p$-forms by making the increasing exterior products of an orthonormal coframe orthonormal. The Hodge star operator is the unique pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ satisfyingFor an increasing multi-index $I$, $\ast e^I$ is the complementary wedge with the sign making $e^I\wedge \ast e^I={\omega }_g$. Swapping the blocks of $p$ and $n-p$ factors introduces $(-1{)}^{p(n-p)}$. ThereforeIn particular $\ast 1={\omega }_g$ and $\ast {\omega }_g=1$.

For compactly supported smooth $f$, the $(n-1)$-form $f\ast \alpha$ has compact support. Stokes theorem and the graded Leibniz rule giveA top form $\tau$ equals $(\ast \tau ){\omega }_g$. Consequently this Hodge integration by parts for one-forms becomes exactlyCompact support of $f$ is enough; $\alpha$ need not itself have compact support.

Define the codifferential on $p$-forms by $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ and the Hodge Laplacian by $\Delta =d\delta +\delta d$. A harmonic differential form is a smooth form in $\ker \Delta$. On functions this is the positive Laplace-Beltrami operator, $\Delta f=\delta df=-{\mathrm{div}}_g{\mathrm{grad}}_{g}f$. This sign convention is required by the product identity; it is the negative of the $\mathrm{div}\mathrm{grad}$ convention also commonly used for the Laplace-Beltrami operator.

For a $p$-form $\beta$, the square of the Hodge star operator and the codifferential formula giveApplying the second formula to $d\beta$ and the first to $\delta \beta$ givesThus Hodge star commutes with the Hodge Laplacian. Since $\ast$ is invertible, $\beta$ is harmonic if and only if $\ast \beta$ is harmonic. This holds without compactness; we have not used the generally false noncompact implication that a harmonic form must be closed and coclosed.

For a smooth function $h$ and one-form $\alpha$, the same graded Leibniz rule gives $\delta (h\alpha )=h\delta \alpha -\ast (dh\wedge \ast \alpha )=h\delta \alpha -⟨dh,\alpha {⟩}_g$. Apply this to $d(f_1{f}_2)=f_{2}d{f}_1+f_{1}d{f}_2$ to obtain the product rule for the positive Laplace-Beltrami operator

The Hodge decomposition theorem states that on a compact oriented boundaryless Riemannian manifold, smooth forms have the $L^2$-orthogonal decompositionand each de Rham cohomology class has a unique harmonic differential form representative. To spell out the last conclusion, a harmonic form is closed and coclosed because $⟨\Delta \eta ,\eta ⟩=\Vert d\eta {\Vert }^2+\Vert \delta \eta {\Vert }^2$. If a closed form decomposes as $h+du+\delta v$, then $d\delta v=0$; integration by parts gives $\Vert \delta v{\Vert }^2=⟨v,d\delta v⟩=0$. Thus it represents $h$. A harmonic exact form has zero norm by adjointness, proving uniqueness. The analytic existence of the decomposition is the stated Hodge decomposition theorem.

Now let $M$ be compact, connected and oriented. A harmonic function satisfies $0=⟨f,\Delta f⟩=\Vert df{\Vert }^2$, so it is constant. The Hodge star operator identifies ${\mathcal{H}}^0(M)$ with ${\mathcal{H}}^n(M)$, hence ${\mathcal{H}}^n(M)=\mathbb{R}{\omega }_g$. A Riemannian metric exists by Question 3 even if none was initially chosen. Using the harmonic representative of each class, we obtain the top de Rham cohomology of a compact connected oriented manifoldThe class is nonzero also directly from Stokes theorem, since ${\int }_M{\omega }_g>0$ while every exact top form has zero integral. Boundarylessness matters: a compact interval has zero first de Rham cohomology.

For a compact connected oriented boundaryless $n$-dimensional smooth manifold, choose a Riemannian metric. Harmonic functions are constant because $⟨f,\Delta f⟩=\Vert df{\Vert }^2$, and Hodge star commutes with the Hodge Laplacian. Thus the Hodge star operator identifies the constant functions with the harmonic differential forms of top degree, exactly $\mathbb{R}{\omega }_g$. The Hodge decomposition theorem identifies these with the top de Rham cohomology. The metric volume form has nonzero class by Stokes theorem. The harmonic-representative theorem is stated in Denis Auroux's Hodge theory lecture.