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At each point, use the inverse metric on the cotangent space. It induces an inner product on $p$-forms byEquivalently, wedges of distinct members of an orthonormal coframe form an orthonormal basis. Choose an orientation to define the ordinary global Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩d{\mathrm{vol}}_g$. It maps $p$-forms to $(n-p)$-forms and obeys ${\ast }^2=(-1{)}^{p(n-p)}$. We use real forms; complex forms are obtained by complex-linear extension. Without an orientation an ordinary global star is not available without twisting the target, so orientation is implicit in this first part.

With the codifferential ${\delta }_p=(-1{)}^{n(p+1)+1}\ast d\ast$ for $p>0$ and ${\delta }_0=0$, define the Hodge Laplacian $\Delta =d\delta +\delta d$. On functions it is the nonnegative-sign Laplace-Beltrami operator, $\Delta f=-\mathrm{div}\mathrm{grad}f$. A harmonic differential form satisfies $\Delta \alpha =0$.

For $\alpha$ of degree $p$, the displayed sign formula and star square giveApplying the identities again, at degrees $p-1$ and $p+1$, proves the Hodge star commutes with the Hodge Laplacian relationSince star is invertible, $\alpha$ is harmonic if and only if $\ast \alpha$ is harmonic. This part does not need compactness; zero-degree and top-degree terms are interpreted as zero when their degrees are outside the range.

For the remaining cohomological statements assume a compact oriented manifold without boundary. The Hodge decomposition theorem gives the $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^p=\ker \Delta$ is finite-dimensional and consists of smooth forms. Formal adjointness gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic differential forms are closed differential forms and coclosed differential forms.

If $\Delta f\ge 0$, formal adjointness and $\Delta 1=0$ give ${\int }_M\Delta f=0$. A continuous nonnegative function with zero integral vanishes, so $\Delta f=0$. Then $\Vert df{\Vert }_{L^2}^2=⟨\Delta f,f⟩=0$. Thus the a function with nonnegative Laplacian on a closed manifold is locally constant assertion isIt is globally constant when $M$ is connected. The PDF does not explicitly include connectedness here; on two disjoint circles one may take different constants, so literal global constancy needs that qualification. Boundaryless is the standard manifold convention in this argument; otherwise boundary conditions are necessary.

For a closed differential form $\alpha$, write its Hodge decomposition as $\alpha =h+d\beta +\delta \gamma$. Since $d\alpha =dh=0$, we have $d\delta \gamma =0$. Formal adjointness yields $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$, so $\alpha =h+d\beta$. Thus $h$ represents its de Rham cohomology class. If two harmonic representatives differ by $d\beta$, their difference $h$ has $\Vert h{\Vert }^2=⟨d\beta ,h⟩=⟨\beta ,\delta h⟩=0$. Therefore

Finally let $h\in G$. A connected Lie group admits a smooth path $h_t$ from the identity to $h$. The maps $R_{h_t}$ give a smooth homotopy between the identity and $R_h$. The homotopy invariance of de Rham cohomology says their pullbacks agree on cohomology. The maps $R_{h_t}$ preserve orientation, since their Jacobian signs cannot change from the identity along the path. As an orientation-preserving isometry, $R_h$ commutes on forms with star, exterior differentiation, codifferentiation and consequently the Hodge Laplacian. Hence $R_h^{\ast }\alpha$ is harmonic and represents the same cohomology class as a harmonic $\alpha$. Uniqueness now proves the harmonic forms are fixed by a connected isometric group action conclusionThe argument also applies componentwise when $M$ is disconnected.