[proofplan] The map $\mathcal{H}^p \to H_{\mathrm{dR}}^p(M)$, $\alpha \mapsto [\alpha]$, is well-defined because every harmonic form is closed (by [Harmonic Iff Closed and Co-closed](/theorems/2744)). Injectivity comes from the fact that a harmonic form which is exact is orthogonal to itself in $L^2$: if $\alpha \in \mathcal{H}^p$ and $\alpha = d\beta$, then pairing with $\alpha$ and using that $\alpha$ is co-closed converts the equation to $\|\alpha\|_g^2 = 0$, forcing $\alpha = 0$. Surjectivity uses the [Hodge Decomposition](/theorems/2745): given a closed form $\eta$, its $\mathcal{H}^p$-component $\alpha$ has the same cohomology class as $\eta$, because the orthogonal complement $\mathcal{H}^{p\perp} = \Delta\Omega^p(M)$ intersects the closed forms in exactly the exact forms (by the [Refined Hodge Decomposition](/theorems/2746)). Combining, $\dim H_{\mathrm{dR}}^p(M) = \dim \mathcal{H}^p$, which is finite by the Hodge Decomposition. [/proofplan]

[step:Verify that $\alpha \mapsto [\alpha]$ is a well-defined linear map $\mathcal{H}^p \to H_{\mathrm{dR}}^p(M)$] Let $(M, g)$ be a compact oriented Riemannian manifold without boundary, and let $\mathcal{H}^p := \{\alpha \in \Omega^p(M) : \Delta\alpha = 0\}$ denote the space of harmonic $p$-forms. By the [Harmonic Iff Closed and Co-closed](/theorems/2744), every $\alpha \in \mathcal{H}^p$ satisfies $d\alpha = 0$, i.e., $\alpha$ is closed. Therefore $\alpha$ defines a de Rham cohomology class \begin{align*} [\alpha] := \alpha + d\Omega^{p-1}(M) \in H_{\mathrm{dR}}^p(M), \end{align*} and the assignment \begin{align*} \Phi : \mathcal{H}^p &\to H_{\mathrm{dR}}^p(M), \\ \alpha &\mapsto [\alpha] \end{align*} is well-defined. Linearity of $\Phi$ is immediate from linearity of the projection $\Omega^p_\mathrm{cl}(M) \to H_{\mathrm{dR}}^p(M)$ from closed forms onto cohomology classes. [/step]

[step:Set up the global $L^2$ inner product to convert exactness statements into orthogonality] Let $\omega_g$ be the Riemannian volume form, $(\cdot, \cdot)_g$ the fibrewise inner product on $\Lambda^p T^*M$, and \begin{align*} \langle\langle \cdot, \cdot \rangle\rangle_g : \Omega^p(M) \times \Omega^p(M) &\to \mathbb{R}, \\ (\eta, \xi) &\mapsto \int_M (\eta, \xi)_g \, \omega_g \end{align*} the global $L^2$ inner product, with norm $\|\eta\|_g$. The formal adjoint identity from the [Co-differential is Formal Adjoint of $d$](/theorems/2742), \begin{align*} \langle\langle d\eta, \xi \rangle\rangle_g = \langle\langle \eta, \delta\xi \rangle\rangle_g \qquad \eta \in \Omega^{p-1}(M), \ \xi \in \Omega^p(M), \tag{A} \end{align*} holds for all smooth forms on the closed manifold $M$ (no compact support hypothesis is needed because $M$ itself is compact). [/step]

[step:Prove $\Phi$ is injective by showing harmonic-and-exact forces zero]Suppose $\alpha \in \mathcal{H}^p$ with $\Phi(\alpha) = 0$, i.e., $\alpha = d\beta$ for some $\beta \in \Omega^{p-1}(M)$. We show $\alpha = 0$. Compute $\|\alpha\|_g^2$ by substituting $\alpha = d\beta$ on one factor: \begin{align*} \|\alpha\|_g^2 = \langle\langle \alpha, \alpha \rangle\rangle_g = \langle\langle \alpha, d\beta \rangle\rangle_g. \end{align*} Apply (A) with $\eta := \beta$ and $\xi := \alpha$, using symmetry of the real $L^2$ pairing to put the right factor in the form $d\eta$: \begin{align*} \langle\langle \alpha, d\beta \rangle\rangle_g = \langle\langle d\beta, \alpha \rangle\rangle_g = \langle\langle \beta, \delta\alpha \rangle\rangle_g. \end{align*} Since $\alpha \in \mathcal{H}^p$, the [Harmonic Iff Closed and Co-closed](/theorems/2744) gives $\delta\alpha = 0$. Hence \begin{align*} \|\alpha\|_g^2 = \langle\langle \beta, 0 \rangle\rangle_g = 0, \end{align*} and since $\|\cdot\|_g$ is positive-definite on $\Omega^p(M)$ (the fibrewise inner product is positive-definite, the volume form is positive, and $M$ is compact, so $\|\eta\|_g = 0$ forces $(\eta, \eta)_g \equiv 0$, hence $\eta \equiv 0$), \begin{align*} \alpha = 0. \end{align*} This proves $\ker\Phi = \{0\}$, so $\Phi$ is injective.[/step]

[guided]Injectivity of a linear map is equivalent to triviality of its kernel, so we suppose $\alpha \in \mathcal{H}^p$ lies in $\ker\Phi$ — i.e., $[\alpha] = 0$ in $H_{\mathrm{dR}}^p(M)$ — and aim to deduce $\alpha = 0$. By definition of the de Rham quotient, $[\alpha] = 0$ means $\alpha = d\beta$ for some primitive $\beta \in \Omega^{p-1}(M)$. How do we extract that $\alpha$ vanishes from this? The standard trick on a compact manifold is to feed $\alpha$ to the $L^2$ norm: showing $\|\alpha\|_g = 0$ forces $\alpha = 0$ pointwise, since $\|\cdot\|_g$ is positive-definite on $\Omega^p(M)$. So we compute $\|\alpha\|_g^2$ and substitute the exact form $d\beta$ on one factor: \begin{align*} \|\alpha\|_g^2 = \langle\langle \alpha, \alpha \rangle\rangle_g = \langle\langle \alpha, d\beta \rangle\rangle_g. \end{align*} Now we have a $d$ inside the pairing — the perfect setup for the formal adjoint identity (A). The strategy is to move $d$ off the right-hand factor (where it is acting on $\beta$) and convert it into $\delta$ acting on $\alpha$ on the left-hand side, because we know $\delta\alpha = 0$. Identity (A) is stated with $d$ on the left factor, so we first use symmetry of the real $L^2$ pairing to put it in that form, then apply (A) with $\eta := \beta$ and $\xi := \alpha$: \begin{align*} \langle\langle \alpha, d\beta \rangle\rangle_g = \langle\langle d\beta, \alpha \rangle\rangle_g = \langle\langle \beta, \delta\alpha \rangle\rangle_g. \end{align*} This is where the harmonicity of $\alpha$ is consumed. By the [Harmonic Iff Closed and Co-closed](/theorems/2744), $\Delta\alpha = 0$ implies both $d\alpha = 0$ and $\delta\alpha = 0$. We use the second: $\delta\alpha = 0$, so the right pairing collapses: \begin{align*} \|\alpha\|_g^2 = \langle\langle \beta, 0 \rangle\rangle_g = 0. \end{align*} It remains to convert $\|\alpha\|_g = 0$ into $\alpha = 0$. The norm $\|\cdot\|_g$ is positive-definite on $\Omega^p(M)$: the fibrewise inner product $(\cdot, \cdot)_g$ is positive-definite on each fibre $\Lambda^p T_x^*M$, the volume form $\omega_g$ is a positive density (since $M$ is oriented), and $M$ is compact so the integral is finite. Hence $\|\eta\|_g = 0$ forces $(\eta, \eta)_g \equiv 0$ pointwise, which forces $\eta \equiv 0$. Applied to $\alpha$: \begin{align*} \alpha = 0. \end{align*} We conclude $\ker\Phi = \{0\}$, so $\Phi$ is injective. Notice that the same computation, applied to the difference of two harmonic forms in the same cohomology class, gives uniqueness of the harmonic representative — but here we only needed the special case "the only harmonic exact form is zero", which is exactly $\ker\Phi = \{0\}$.[/guided]

[step:Prove $\Phi$ is surjective by extracting the harmonic component of a closed form]Let $\eta \in \Omega^p(M)$ be closed, $d\eta = 0$. We must find $\alpha \in \mathcal{H}^p$ with $[\alpha] = [\eta]$, i.e. $\alpha - \eta = d\gamma$ for some $\gamma \in \Omega^{p-1}(M)$. By the [Refined Hodge Decomposition](/theorems/2746), \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus d\Omega^{p-1}(M) \oplus \delta\Omega^{p+1}(M), \tag{D2} \end{align*} as an orthogonal direct sum in $\langle\langle \cdot, \cdot \rangle\rangle_g$. Decompose $\eta$ accordingly: \begin{align*} \eta = \alpha + d\gamma + \delta\rho, \end{align*} with $\alpha \in \mathcal{H}^p$, $\gamma \in \Omega^{p-1}(M)$, $\rho \in \Omega^{p+1}(M)$. We claim $\delta\rho = 0$. Apply $d$ to both sides of the decomposition. Since $d\eta = 0$, $d\alpha = 0$ (harmonic forms are closed by [Harmonic Iff Closed and Co-closed](/theorems/2744)), and $d \circ d = 0$ kills $d(d\gamma) = 0$, \begin{align*} 0 = d\eta = d\alpha + d^2\gamma + d\delta\rho = 0 + 0 + d\delta\rho = d\delta\rho. \end{align*} Therefore $d\delta\rho = 0$. Pair $\delta\rho$ with itself in $L^2$ and use (A) at degree $p \to p+1$ with $\eta_A := \rho \in \Omega^{p+1}(M)$ and the complementary identity to flip $\delta$ to $d$: \begin{align*} \|\delta\rho\|_g^2 = \langle\langle \delta\rho, \delta\rho \rangle\rangle_g = \langle\langle \rho, d\delta\rho \rangle\rangle_g = \langle\langle \rho, 0 \rangle\rangle_g = 0, \end{align*} where the second equality uses (A) read in the form \begin{align*} \langle\langle \delta\sigma, \tau \rangle\rangle_g = \langle\langle \sigma, d\tau \rangle\rangle_g \qquad \sigma \in \Omega^{p+1}(M), \ \tau \in \Omega^p(M), \tag{A'} \end{align*} which follows from (A) by symmetry of the pairing (set $\eta := \tau$, $\xi := \sigma$ in (A) and use symmetry to obtain (A')). By positive-definiteness of $\|\cdot\|_g$, $\delta\rho = 0$. Therefore \begin{align*} \eta = \alpha + d\gamma, \end{align*} so $\alpha - \eta = -d\gamma \in d\Omega^{p-1}(M)$, i.e., $[\alpha] = [\eta]$ in $H_{\mathrm{dR}}^p(M)$. Hence $\Phi(\alpha) = [\eta]$, and $\Phi$ is surjective.[/step]

[guided]Surjectivity is where the Hodge decomposition does the heavy lifting. We must show that every cohomology class $[\eta] \in H_{\mathrm{dR}}^p(M)$ — represented by a closed form $\eta \in \Omega^p(M)$ with $d\eta = 0$ — contains a harmonic representative. Concretely, we want $\alpha \in \mathcal{H}^p$ with $\alpha - \eta \in d\Omega^{p-1}(M)$, i.e. $\alpha - \eta = d\gamma$ for some $\gamma \in \Omega^{p-1}(M)$. The natural candidate for $\alpha$ is the $\mathcal{H}^p$-projection of $\eta$. Why does such a projection exist? Because the [Refined Hodge Decomposition](/theorems/2746) splits $\Omega^p(M)$ into three orthogonal pieces: \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus d\Omega^{p-1}(M) \oplus \delta\Omega^{p+1}(M), \tag{D2} \end{align*} as an orthogonal direct sum in $\langle\langle \cdot, \cdot \rangle\rangle_g$. Writing $\eta$ in this decomposition, \begin{align*} \eta = \alpha + d\gamma + \delta\rho, \end{align*} with $\alpha \in \mathcal{H}^p$, $\gamma \in \Omega^{p-1}(M)$, $\rho \in \Omega^{p+1}(M)$. The harmonic and the exact pieces are exactly the data we want — $\alpha$ is the harmonic candidate and $d\gamma$ is the exact difference. The obstruction is the third piece $\delta\rho$: we need it to vanish, otherwise $\alpha - \eta$ contains a co-exact term that is not exact. So the entire surjectivity argument reduces to one claim: **$\delta\rho = 0$**. Where will $\delta\rho = 0$ come from? We have not yet used the hypothesis $d\eta = 0$, so that closedness must be the source. Applying $d$ to both sides of the decomposition and tracking each term: - $d\eta = 0$ by hypothesis. - $d\alpha = 0$ because $\alpha \in \mathcal{H}^p$ is harmonic, hence closed by [Harmonic Iff Closed and Co-closed](/theorems/2744). - $d(d\gamma) = d^2\gamma = 0$ because $d \circ d = 0$. This leaves only the $\delta\rho$-term: \begin{align*} 0 = d\eta = d\alpha + d^2\gamma + d\delta\rho = 0 + 0 + d\delta\rho = d\delta\rho. \end{align*} So $d\delta\rho = 0$. We have not yet shown $\delta\rho$ itself vanishes — only that its differential does. How to upgrade $d\delta\rho = 0$ to $\delta\rho = 0$? The same trick as in injectivity: pair against itself in $L^2$ and flip the operator across the pairing. We need the formal adjoint identity in the form \begin{align*} \langle\langle \delta\sigma, \tau \rangle\rangle_g = \langle\langle \sigma, d\tau \rangle\rangle_g \qquad \sigma \in \Omega^{p+1}(M), \ \tau \in \Omega^p(M), \tag{A'} \end{align*} which is just (A) read in the opposite direction: setting $\eta := \tau$ and $\xi := \sigma$ in (A) gives $\langle\langle d\tau, \sigma \rangle\rangle_g = \langle\langle \tau, \delta\sigma \rangle\rangle_g$, and symmetry of the real $L^2$ pairing on both sides yields (A'). Apply (A') with $\sigma := \rho$ and $\tau := \delta\rho$: \begin{align*} \|\delta\rho\|_g^2 = \langle\langle \delta\rho, \delta\rho \rangle\rangle_g = \langle\langle \rho, d\delta\rho \rangle\rangle_g = \langle\langle \rho, 0 \rangle\rangle_g = 0, \end{align*} using $d\delta\rho = 0$ from above. By positive-definiteness of $\|\cdot\|_g$ (same justification as in injectivity: pointwise inner product positive, volume form positive, $M$ compact), we conclude $\delta\rho = 0$. With the obstruction killed, the decomposition collapses to \begin{align*} \eta = \alpha + d\gamma, \end{align*} so $\alpha - \eta = -d\gamma \in d\Omega^{p-1}(M)$, hence $[\alpha] = [\eta]$ in $H_{\mathrm{dR}}^p(M)$. Therefore $\Phi(\alpha) = [\alpha] = [\eta]$, and $\Phi$ is surjective. This is the essence of "the harmonic representative is unique and canonical" — combined with injectivity, $\Phi$ being a bijection means every cohomology class has a *unique* harmonic representative, so the harmonic representative is the canonical representative.[/guided]

[step:Conclude $\Phi$ is an isomorphism and read off the dimension equality] We have shown $\Phi : \mathcal{H}^p \to H_{\mathrm{dR}}^p(M)$ is linear, injective, and surjective. Hence $\Phi$ is a linear isomorphism. By the [Hodge Decomposition](/theorems/2745), $\mathcal{H}^p$ is finite-dimensional. Therefore \begin{align*} \dim H_{\mathrm{dR}}^p(M) = \dim \mathcal{H}^p < \infty. \end{align*} This completes the proof. [/step]