[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 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 argument), $\mathcal{H}^p$ is finite-dimensional.[/step]

[guided]The structure of this argument is "$\mathcal{H}^p$ is a closed subspace of an infinite-dimensional Hilbert space whose closed unit ball is precompact, hence finite-dimensional". The Compactness Theorem provides the precompactness: a uniform bound on $\|\alpha_n\|_g$ and $\|\Delta\alpha_n\|_g$ gives an $L^2$-Cauchy subsequence. On harmonic forms the second bound is automatic, so the precompactness reduces to "bounded sequences of harmonic forms have $L^2$-convergent subsequences". This is Rellich-type compactness for the operator $\Delta$. The Regularity Theorem then upgrades the $L^2$-limit to a smooth harmonic form, closing the loop. The precise statement of the functional-analytic input is: if $X$ is a normed vector space and the closed unit ball $\overline{B}_X(0,1)$ is sequentially compact, then $\dim X < \infty$. This is the contrapositive of Riesz's lemma, which produces in any infinite-dimensional normed space an infinite sequence in the unit ball with pairwise distances $\ge 1/2$ — such a sequence has no Cauchy subsequence.[/guided]

[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 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).[/step]

[guided]The non-trivial input here is that $\Delta\Omega^p(M)$ is closed in $\Omega^p(M)$ for the $L^2$ topology — this is the content of the existence theory for the Hodge Laplacian and is what allows us to split off a harmonic part by orthogonal projection. Concretely: given $\beta \in \Omega^p(M)$, we want to find $\gamma \in \Omega^p(M)$ with $\Delta\gamma = \beta - \alpha$ for some $\alpha \in \mathcal{H}^p$. The form $\alpha$ must be the orthogonal projection of $\beta$ onto $\mathcal{H}^p$, which exists and is unique because $\mathcal{H}^p$ is finite-dimensional. The remaining task is then to solve $\Delta\gamma = \beta - \alpha$, which is solvable precisely because $\beta - \alpha \in \mathcal{H}^{p\perp} = \Delta\Omega^p(M)$. The closedness of $\Delta\Omega^p(M)$ is a consequence of the Compactness Theorem combined with the Regularity Theorem: if $\Delta\gamma_n \to \tau$ in $L^2$ and we replace $\gamma_n$ by its $\mathcal{H}^{p\perp}$-component (which preserves $\Delta\gamma_n$), one obtains uniform $L^2$-control of $\gamma_n$, then a Cauchy subsequence, then a smooth limit $\gamma$ with $\Delta\gamma = \tau$. The full argument requires the Hahn-Banach Theorem, the Regularity Theorem, and the Compactness Theorem assembled together; we state it here as the closedness assertion and refer to those theorems for the full chain.[/guided]

[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]