[proofplan] This proof is a sketch of the main steps; the full functional-analytic argument requires elliptic regularity for the Hodge Laplacian on a compact manifold (the [Regularity Theorem](/theorems/2750)) and a compactness statement for sequences of smooth forms with bounded $L^2$ norm and bounded Laplacian (the [Compactness Theorem](/theorems/2748)). We give the strategy in full and indicate where each black box is invoked. The proof has two main parts: (1) finite-dimensionality of $\mathcal{H}^p$, which follows because the unit ball in $\mathcal{H}^p$ satisfies the hypotheses of the [Compactness Theorem](/theorems/2748) and a finite-dimensional normed space is the only one whose closed unit ball is sequentially compact; and (2) the orthogonal direct sum $\Omega^p(M) = \mathcal{H}^p \oplus \Delta\Omega^p(M)$, which follows from the orthogonal complement characterisation $\mathcal{H}^p = (\Delta\Omega^p(M))^\perp$ together with the closedness of $\Delta\Omega^p(M)$ in $\Omega^p(M)$ for the $L^2$ topology, the latter being the content of the existence half of the regularity-plus-compactness package. [/proofplan] [step:Set up the global $L^2$ inner product and the orthogonality of $\mathcal{H}^p$ with $\Delta\Omega^p(M)$] Let $(M, g)$ be a compact oriented Riemannian manifold, $\omega_g$ its 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. Since $M$ is compact and the integrand is continuous, this pairing is finite on all of $\Omega^p(M) \times \Omega^p(M)$. The associated norm is $\|\eta\|_g := \langle\langle \eta, \eta \rangle\rangle_g^{1/2}$, and the space of harmonic $p$-forms is \begin{align*} \mathcal{H}^p := \{\alpha \in \Omega^p(M) : \Delta\alpha = 0\}. \end{align*} We first verify orthogonality: for any $\alpha \in \mathcal{H}^p$ and any $\beta \in \Omega^p(M)$, \begin{align*} \langle\langle \alpha, \Delta\beta \rangle\rangle_g \overset{(*)}{=} \langle\langle \Delta\alpha, \beta \rangle\rangle_g = \langle\langle 0, \beta \rangle\rangle_g = 0, \end{align*} where $(*)$ uses the [Laplacian is Formally Self-Adjoint](/theorems/2743) identity, whose hypotheses (compactness of $M$, smooth forms) are satisfied. So $\mathcal{H}^p \subseteq (\Delta\Omega^p(M))^\perp$ in the $L^2$ pairing on $\Omega^p(M)$. Conversely, if $\alpha \in (\Delta\Omega^p(M))^\perp$, then for every $\beta \in \Omega^p(M)$, \begin{align*} 0 = \langle\langle \alpha, \Delta\beta \rangle\rangle_g = \langle\langle \Delta\alpha, \beta \rangle\rangle_g. \end{align*} Specialising to $\beta := \Delta\alpha$ gives $\|\Delta\alpha\|_g^2 = 0$, hence $\Delta\alpha = 0$ by positive-definiteness of $\|\cdot\|_g$ on smooth forms over a compact manifold. So \begin{align*} \mathcal{H}^p = (\Delta\Omega^p(M))^\perp, \tag{O} \end{align*} where the orthogonal complement is taken inside $\Omega^p(M)$ with respect to $\langle\langle \cdot, \cdot \rangle\rangle_g$. [/step] [step:Show $\mathcal{H}^p$ is finite-dimensional via the Compactness Theorem] We show that the closed unit ball $B := \{\alpha \in \mathcal{H}^p : \|\alpha\|_g \le 1\}$ of $\mathcal{H}^p$ is sequentially compact in the $L^2$-norm topology, hence that $\mathcal{H}^p$ — as a normed space whose closed unit ball is sequentially compact — is finite-dimensional. Let $(\alpha_n) \subseteq B$ be a sequence. Then $\|\alpha_n\|_g \le 1$ and $\Delta\alpha_n = 0$, so $\|\Delta\alpha_n\|_g = 0 \le 1$. The sequence $(\alpha_n)$ thus satisfies the hypotheses of the [Compactness Theorem](/theorems/2748) with $C = 1$: it is a sequence of smooth $p$-forms on $(M, g)$ with uniformly bounded $L^2$-norm and uniformly bounded Laplacian. Theorem 2748 produces a subsequence $(\alpha_{n_k})$ that is Cauchy in the $L^2$-norm. Let $\alpha$ denote its $L^2$-limit. A priori, $\alpha$ is only an element of the $L^2$-completion of $\Omega^p(M)$, not necessarily a smooth form. We must promote $\alpha$ to a smooth harmonic representative. First, $\alpha$ is a *weak* solution of $\Delta\alpha = 0$. For any $\beta \in \Omega^p(M)$, formal self-adjointness ([Laplacian is Formally Self-Adjoint](/theorems/2743)) gives $\langle\langle \alpha_{n_k}, \Delta\beta \rangle\rangle_g = \langle\langle \Delta\alpha_{n_k}, \beta \rangle\rangle_g = 0$ for each $k$. Cauchy-Schwarz with $\alpha_{n_k} \to \alpha$ in $L^2$-norm gives \begin{align*} |\langle\langle \alpha, \Delta\beta \rangle\rangle_g - \langle\langle \alpha_{n_k}, \Delta\beta \rangle\rangle_g| \le \|\alpha - \alpha_{n_k}\|_g \, \|\Delta\beta\|_g \to 0, \end{align*} so $\langle\langle \alpha, \Delta\beta \rangle\rangle_g = 0$ for every $\beta \in \Omega^p(M)$. Second, we *quote* the [Regularity Theorem](/theorems/2750) for the Hodge Laplacian: a weak $L^2$-solution of $\Delta\alpha = 0$ on a compact Riemannian manifold admits a smooth representative. We invoke this as a stated theorem from elliptic PDE — it is the standard interior regularity result for $\Delta$ and is not proved here. Replacing $\alpha$ by its smooth representative gives $\alpha \in \mathcal{H}^p$. Finally, $\|\alpha\|_g \le \liminf_k \|\alpha_{n_k}\|_g \le 1$ by lower semicontinuity of the $L^2$-norm, so $\alpha \in B$. Thus every sequence in $B$ has a subsequence converging in $L^2$-norm to an element of $B$, i.e. $B$ is sequentially compact. By a standard theorem in functional analysis (a normed space whose closed unit ball is compact in the norm topology is finite-dimensional — [Riesz's lemma](/page/Riesz's%20Lemma) argument), $\mathcal{H}^p$ is finite-dimensional. [guided] The structure of the argument is: $\mathcal{H}^p$ is a normed space whose closed unit ball is sequentially compact in the $L^2$ topology, and a normed space with this property is finite-dimensional. The reason this works is that harmonic forms automatically satisfy the bounds needed by the [Compactness Theorem](/theorems/2748), so a unit-ball sequence in $\mathcal{H}^p$ has an $L^2$-Cauchy subsequence. The remaining task is to verify that the $L^2$-limit lies back in $\mathcal{H}^p$ — that is where the [Regularity Theorem](/theorems/2750) enters. Set \begin{align*} B := \{\alpha \in \mathcal{H}^p : \|\alpha\|_g \le 1\}, \end{align*} the closed unit ball of $(\mathcal{H}^p, \|\cdot\|_g)$. We aim to show $B$ is sequentially compact and conclude $\dim \mathcal{H}^p < \infty$. *Extracting an $L^2$-Cauchy subsequence.* Fix any sequence $(\alpha_n) \subseteq B$. Then $\|\alpha_n\|_g \le 1$ by membership in $B$, and $\Delta\alpha_n = 0$ (by definition of $\mathcal{H}^p$), which forces $\|\Delta\alpha_n\|_g = 0 \le 1$. The hypotheses of the [Compactness Theorem](/theorems/2748) are: a sequence of smooth $p$-forms on $(M,g)$ with $\|\alpha_n\|_g \le C$ and $\|\Delta\alpha_n\|_g \le C$ for some $C$; we just verified them with $C = 1$. The conclusion is the existence of an $L^2$-Cauchy subsequence $(\alpha_{n_k})$. Why the second bound is "free" here is the key observation — for general $p$-forms one would have to control both norms separately, but $\mathcal{H}^p$ is precisely the kernel of $\Delta$, so the second bound holds with constant $0$. *Identifying the limit as a weak harmonic form.* Let $\alpha$ be the $L^2$-limit of $(\alpha_{n_k})$. A priori $\alpha$ lies in the $L^2$-completion $\widetilde{\Omega}^p$ and need not be smooth. To extract a smooth harmonic representative we first show $\alpha$ is a weak solution of $\Delta\alpha = 0$, meaning $\langle\langle \alpha, \Delta\beta \rangle\rangle_g = 0$ for every test form $\beta \in \Omega^p(M)$. For each $k$, formal self-adjointness ([Laplacian is Formally Self-Adjoint](/theorems/2743), whose hypotheses — compactness of $M$ and smoothness of the forms — hold) gives \begin{align*} \langle\langle \alpha_{n_k}, \Delta\beta \rangle\rangle_g = \langle\langle \Delta\alpha_{n_k}, \beta \rangle\rangle_g = \langle\langle 0, \beta \rangle\rangle_g = 0. \end{align*} We now pass to the limit using Cauchy-Schwarz on the bilinear pairing: \begin{align*} |\langle\langle \alpha, \Delta\beta \rangle\rangle_g - \langle\langle \alpha_{n_k}, \Delta\beta \rangle\rangle_g| = |\langle\langle \alpha - \alpha_{n_k}, \Delta\beta \rangle\rangle_g| \le \|\alpha - \alpha_{n_k}\|_g \, \|\Delta\beta\|_g \to 0. \end{align*} Combining the two displays, \begin{align*} \langle\langle \alpha, \Delta\beta \rangle\rangle_g = 0 \quad \text{for every } \beta \in \Omega^p(M), \end{align*} which is exactly the weak formulation of $\Delta\alpha = 0$. *Promoting to a smooth solution via regularity.* This is where elliptic regularity does the work. We invoke the [Regularity Theorem](/theorems/2750) for the Hodge Laplacian: a weak $L^2$-solution of $\Delta\alpha = 0$ on a compact Riemannian manifold has a smooth representative. (We are quoting this as a black box from elliptic PDE — proving it would require Sobolev spaces, Gårding's inequality, and a bootstrap argument we do not reproduce.) Replacing $\alpha$ by its smooth representative gives $\alpha \in \Omega^p(M)$ with $\Delta\alpha = 0$, i.e. $\alpha \in \mathcal{H}^p$. *Verifying $\alpha \in B$.* By [lower semicontinuity of the norm](/theorems/215) under $L^2$-convergence, \begin{align*} \|\alpha\|_g \le \liminf_{k \to \infty} \|\alpha_{n_k}\|_g \le 1, \end{align*} so $\alpha \in B$. *Concluding finite-dimensionality.* Every sequence in $B$ has a subsequence converging in the $L^2$-norm to an element of $B$, so $B$ is sequentially compact. We now invoke the functional-analytic input: if $X$ is a [normed vector space](/page/Normed%20Vector%20Space) whose closed unit ball $\overline{B}_X(0,1)$ is sequentially compact, then $\dim X < \infty$. This is the contrapositive of [Riesz's lemma](/theorems/1222) — in any infinite-dimensional normed space one constructs an infinite sequence in the unit ball with pairwise distances $\ge 1/2$, and such a sequence has no Cauchy subsequence and hence no convergent subsequence. Applying this to $X = \mathcal{H}^p$ (with the $L^2$-norm), we conclude \begin{align*} \dim \mathcal{H}^p < \infty. \end{align*} [/guided] [/step] [step:Identify $\Delta\Omega^p(M)$ as the orthogonal complement of $\mathcal{H}^p$ via closedness and (O)] We now show \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus \Delta\Omega^p(M) \tag{D} \end{align*} as an orthogonal direct sum in $\langle\langle \cdot, \cdot \rangle\rangle_g$. Orthogonality of the two summands is identity (O) from the first step. We must show that every $\beta \in \Omega^p(M)$ admits a unique decomposition $\beta = \alpha + \Delta\gamma$ with $\alpha \in \mathcal{H}^p$ and $\gamma \in \Omega^p(M)$. *The pre-Hilbert obstruction.* The space $\Omega^p(M)$ with $\langle\langle\cdot,\cdot\rangle\rangle_g$ is *pre-Hilbert* but not Hilbert: it is dense in but not equal to its $L^2$-completion. Standard orthogonal-decomposition theorems require Hilbert completeness, so we cannot apply them directly to $\Omega^p(M)$. We work in the Hilbert completion $\widetilde{\Omega}^p$ and use elliptic regularity to descend back. Let $\widetilde{\Omega}^p$ denote the $L^2$-completion of $\Omega^p(M)$. Inside $\widetilde{\Omega}^p$, $\mathcal{H}^p$ is finite-dimensional (Step 2) and hence closed, so the standard Hilbert-space [orthogonal decomposition](/theorems/436) gives $\widetilde{\Omega}^p = \mathcal{H}^p \oplus (\mathcal{H}^p)^\perp$. For any smooth $\beta \in \Omega^p(M)$, the projection $\alpha \in \mathcal{H}^p$ is given by an explicit linear-algebra formula in any $L^2$-orthonormal basis of $\mathcal{H}^p$ and is automatically smooth, so $\beta^\perp := \beta - \alpha$ is also smooth. [claim:Poincaré-type lower bound for $\Delta$ on $(\mathcal{H}^p)^\perp$] There exists $c > 0$ such that for all $\eta \in \Omega^p(M) \cap (\mathcal{H}^p)^\perp$, \begin{align*} \|\Delta\eta\|_g \ge c\, \|\eta\|_g. \tag{P} \end{align*} [/claim] [proof] Suppose (P) fails. Then there exists $(\eta_n) \subseteq \Omega^p(M) \cap (\mathcal{H}^p)^\perp$ with $\|\eta_n\|_g = 1$ and $\|\Delta\eta_n\|_g \to 0$. By the [Compactness Theorem](/theorems/2748), $(\eta_n)$ has an $L^2$-Cauchy subsequence $(\eta_{n_k})$ with $L^2$-limit $\eta_\infty \in \widetilde{\Omega}^p$. For any $\beta \in \Omega^p(M)$, formal self-adjointness gives \begin{align*} \langle\langle \eta_\infty, \Delta\beta \rangle\rangle_g = \lim_k \langle\langle \eta_{n_k}, \Delta\beta \rangle\rangle_g = \lim_k \langle\langle \Delta\eta_{n_k}, \beta \rangle\rangle_g, \end{align*} bounded by $\|\Delta\eta_{n_k}\|_g \|\beta\|_g \to 0$. So $\eta_\infty$ is a weak solution of $\Delta\eta_\infty = 0$, and the [Regularity Theorem](/theorems/2750) provides a smooth representative, giving $\eta_\infty \in \mathcal{H}^p$. But $(\mathcal{H}^p)^\perp$ is closed in $\widetilde{\Omega}^p$ and contains each $\eta_{n_k}$, so $\eta_\infty \in (\mathcal{H}^p)^\perp$. Therefore $\eta_\infty \in \mathcal{H}^p \cap (\mathcal{H}^p)^\perp = \{0\}$, i.e. $\eta_\infty = 0$. But $\|\eta_\infty\|_g = \lim_k \|\eta_{n_k}\|_g = 1$, a contradiction. [/proof] *Closedness of $\Delta\Omega^p(M)$ in $L^2$.* Bound (P) makes $\Delta : \Omega^p(M) \cap (\mathcal{H}^p)^\perp \to \Delta\Omega^p(M)$ bounded below. Let $(\Delta\gamma_n) \subseteq \Delta\Omega^p(M)$ be $L^2$-Cauchy with limit $\tau \in \widetilde{\Omega}^p$. Replacing each $\gamma_n$ by its $(\mathcal{H}^p)^\perp$-component (which preserves $\Delta\gamma_n$ since $\Delta\mathcal{H}^p = 0$), bound (P) gives $\|\gamma_n - \gamma_m\|_g \le c^{-1}\|\Delta\gamma_n - \Delta\gamma_m\|_g$, so $(\gamma_n)$ is also $L^2$-Cauchy with some limit $\gamma_\infty$. The same weak-solution + regularity argument yields a smooth $\gamma \in \Omega^p(M)$ with $\Delta\gamma = \tau$, so $\tau \in \Delta\Omega^p(M)$. *Identifying the orthogonal complement.* Inside $\widetilde{\Omega}^p$, $\Delta\Omega^p(M)$ is closed by the previous paragraph. The Hilbert-space identity $(V^\perp)^\perp = V$ for closed $V$, combined with (O), gives $(\mathcal{H}^p)^\perp = \Delta\Omega^p(M)$. Restricting to the smooth level: for $\beta \in \Omega^p(M)$, $\beta = \alpha + \beta^\perp$ with $\alpha \in \mathcal{H}^p$ and $\beta^\perp \in (\mathcal{H}^p)^\perp = \Delta\Omega^p(M)$, both smooth. Writing $\beta^\perp = \Delta\gamma$ for some $\gamma \in \Omega^p(M)$ gives (D). [guided] We want to show \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus \Delta\Omega^p(M) \tag{D} \end{align*} as an orthogonal direct sum in $\langle\langle\cdot,\cdot\rangle\rangle_g$. Orthogonality is identity (O) from Step 1. Existence of the decomposition is the substantive content: every $\beta \in \Omega^p(M)$ should split as $\alpha + \Delta\gamma$ with $\alpha \in \mathcal{H}^p$ and $\gamma \in \Omega^p(M)$. *Why we cannot apply the Hilbert [projection theorem](/theorems/1985) directly.* The pre-[Hilbert space](/page/Hilbert%20Space) $(\Omega^p(M), \langle\langle\cdot,\cdot\rangle\rangle_g)$ is dense in but not equal to its $L^2$-completion $\widetilde{\Omega}^p$. Standard orthogonal-decomposition theorems require Hilbert completeness, so they do not apply to the smooth space directly. The strategy is therefore: pass to $\widetilde{\Omega}^p$, do the projection there, and then descend back to smooth forms using elliptic regularity. *Projecting onto $\mathcal{H}^p$ inside $\widetilde{\Omega}^p$.* By Step 2, $\mathcal{H}^p$ is finite-dimensional and hence closed in $\widetilde{\Omega}^p$, so the Hilbert [projection theorem](/theorems/1985) yields \begin{align*} \widetilde{\Omega}^p = \mathcal{H}^p \oplus (\mathcal{H}^p)^\perp. \end{align*} For $\beta \in \Omega^p(M)$, the projection $\alpha \in \mathcal{H}^p$ is given by the explicit formula $\alpha = \sum_{i=1}^{\dim\mathcal{H}^p} \langle\langle \beta, e_i \rangle\rangle_g \, e_i$ in any $L^2$-orthonormal basis $(e_i)$ of $\mathcal{H}^p$. Since each $e_i$ is smooth, $\alpha$ is smooth, hence so is $\beta^\perp := \beta - \alpha \in (\mathcal{H}^p)^\perp$. *The Poincaré-type lower bound (P) for $\Delta$ on $(\mathcal{H}^p)^\perp$.* The next ingredient is the following claim: there is $c > 0$ such that \begin{align*} \|\Delta\eta\|_g \ge c\, \|\eta\|_g \quad \text{for all } \eta \in \Omega^p(M) \cap (\mathcal{H}^p)^\perp. \tag{P} \end{align*} Why do we expect such a bound? On $\mathcal{H}^p$ the Laplacian is zero, so the bound must fail there; by removing the kernel and looking at $(\mathcal{H}^p)^\perp$ we expect a spectral gap. We argue by contradiction: suppose (P) fails, so there exists $(\eta_n) \subseteq \Omega^p(M) \cap (\mathcal{H}^p)^\perp$ with $\|\eta_n\|_g = 1$ and $\|\Delta\eta_n\|_g \to 0$. Since both $\|\eta_n\|_g$ and $\|\Delta\eta_n\|_g$ are uniformly bounded, the [Compactness Theorem](/theorems/2748) supplies an $L^2$-Cauchy subsequence $(\eta_{n_k})$ with limit $\eta_\infty \in \widetilde{\Omega}^p$. We test against $\Delta\beta$ for $\beta \in \Omega^p(M)$ using formal self-adjointness ([Laplacian is Formally Self-Adjoint](/theorems/2743)): \begin{align*} \langle\langle \eta_\infty, \Delta\beta \rangle\rangle_g = \lim_{k\to\infty} \langle\langle \eta_{n_k}, \Delta\beta \rangle\rangle_g = \lim_{k\to\infty} \langle\langle \Delta\eta_{n_k}, \beta \rangle\rangle_g. \end{align*} Cauchy-Schwarz bounds the right-hand side: \begin{align*} |\langle\langle \Delta\eta_{n_k}, \beta \rangle\rangle_g| \le \|\Delta\eta_{n_k}\|_g \, \|\beta\|_g \to 0. \end{align*} So $\langle\langle \eta_\infty, \Delta\beta \rangle\rangle_g = 0$ for every $\beta$, i.e. $\eta_\infty$ is a weak solution of $\Delta\eta_\infty = 0$. Applying the [Regularity Theorem](/theorems/2750) — whose hypotheses (compact Riemannian manifold, weak $L^2$-solution) are met — gives a smooth representative, so $\eta_\infty \in \mathcal{H}^p$. But $(\mathcal{H}^p)^\perp$ is closed in $\widetilde{\Omega}^p$ (the orthogonal complement of any subset is automatically closed), and each $\eta_{n_k} \in (\mathcal{H}^p)^\perp$, so $\eta_\infty \in (\mathcal{H}^p)^\perp$. Therefore \begin{align*} \eta_\infty \in \mathcal{H}^p \cap (\mathcal{H}^p)^\perp = \{0\}, \end{align*} forcing $\eta_\infty = 0$. On the other hand, by continuity of the norm under $L^2$-convergence, \begin{align*} \|\eta_\infty\|_g = \lim_{k\to\infty} \|\eta_{n_k}\|_g = 1, \end{align*} contradicting $\eta_\infty = 0$. Thus (P) holds. *Closedness of $\Delta\Omega^p(M)$ in the $L^2$-topology.* This is what (P) buys us. Suppose $(\Delta\gamma_n) \subseteq \Delta\Omega^p(M)$ is $L^2$-Cauchy with limit $\tau \in \widetilde{\Omega}^p$. We may replace each $\gamma_n$ by its $(\mathcal{H}^p)^\perp$-component without changing $\Delta\gamma_n$, since $\Delta$ annihilates $\mathcal{H}^p$. Apply (P) to the difference $\gamma_n - \gamma_m \in \Omega^p(M) \cap (\mathcal{H}^p)^\perp$: \begin{align*} \|\gamma_n - \gamma_m\|_g \le c^{-1} \|\Delta\gamma_n - \Delta\gamma_m\|_g \to 0 \quad \text{as } n, m \to \infty. \end{align*} So $(\gamma_n)$ is also $L^2$-Cauchy, with some limit $\gamma_\infty \in \widetilde{\Omega}^p$. The same weak-solution + regularity loop as above promotes $\gamma_\infty$ to a smooth $\gamma \in \Omega^p(M)$ with $\Delta\gamma = \tau$, so $\tau \in \Delta\Omega^p(M)$ and $\Delta\Omega^p(M)$ is closed. The closedness is the existence half of the Hodge story: we can solve $\Delta\gamma = \tau$ for any $\tau$ in the closure. *Identifying the orthogonal complement and finishing (D).* Inside the [Hilbert space](/page/Hilbert%20Space) $\widetilde{\Omega}^p$, $\Delta\Omega^p(M)$ is now a closed subspace, so the standard Hilbert identity $(V^\perp)^\perp = V$ for closed $V$ applies. Combined with (O) — which states $\mathcal{H}^p = (\Delta\Omega^p(M))^\perp$ — we get \begin{align*} (\mathcal{H}^p)^\perp = ((\Delta\Omega^p(M))^\perp)^\perp = \Delta\Omega^p(M). \end{align*} Now restrict to the smooth level: any $\beta \in \Omega^p(M)$ writes as $\beta = \alpha + \beta^\perp$ with $\alpha \in \mathcal{H}^p$ and $\beta^\perp \in (\mathcal{H}^p)^\perp = \Delta\Omega^p(M)$, both smooth. Choosing $\gamma \in \Omega^p(M)$ with $\Delta\gamma = \beta^\perp$ gives \begin{align*} \beta = \alpha + \Delta\gamma, \quad \alpha \in \mathcal{H}^p, \ \gamma \in \Omega^p(M), \end{align*} which is exactly (D). Uniqueness of the decomposition follows from $\mathcal{H}^p \cap \Delta\Omega^p(M) = \{0\}$, itself a consequence of (O): if $\alpha \in \mathcal{H}^p$ and $\alpha = \Delta\gamma$, then $\|\alpha\|_g^2 = \langle\langle \alpha, \Delta\gamma \rangle\rangle_g = 0$ by (O), so $\alpha = 0$. [/guided] [/step] [step:Combine finite-dimensionality and the orthogonal direct sum to conclude] We have established: 1. $\mathcal{H}^p$ is finite-dimensional (Step 2). 2. $\Omega^p(M) = \mathcal{H}^p \oplus \Delta\Omega^p(M)$ as an orthogonal direct sum in $\langle\langle \cdot, \cdot \rangle\rangle_g$ (Step 3). These are exactly the two assertions in the statement of the [Hodge Decomposition](/theorems/2745). The proof is complete. [/step]