[proofplan] The two implications go in opposite directions and use different inputs. The implication $d\alpha = 0$ and $\delta\alpha = 0 \Rightarrow \Delta\alpha = 0$ is a one-line algebraic consequence of the definition $\Delta = d\delta + \delta d$. The converse $\Delta\alpha = 0 \Rightarrow d\alpha = 0$ and $\delta\alpha = 0$ requires compactness of $M$: pairing $\alpha$ against $\Delta\alpha$ in the global $L^2$ inner product and using the fact that $\delta$ is the formal $L^2$ adjoint of $d$ converts the equation into $\|d\alpha\|_g^2 + \|\delta\alpha\|_g^2 = 0$, from which both summands vanish. Compactness enters because it removes the boundary terms in the integration-by-parts that defines $\delta$ as the adjoint of $d$, and because it gives meaning to the $L^2$ pairing on smooth forms without further integrability hypotheses. [/proofplan]

[step:Set up the global $L^2$ pairing on a compact oriented Riemannian manifold]Let $(M, g)$ be a compact oriented Riemannian manifold of dimension $n$ without boundary, and write $\omega_g$ for its Riemannian volume form. The fibrewise inner product on $\Lambda^p T^*M$ induced by $g$ is denoted $(\cdot, \cdot)_g$, and the global $L^2$ pairing on smooth $p$-forms is \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*} Since $M$ is compact and the integrand $(\eta, \xi)_g$ is a continuous function on $M$ for any smooth $\eta, \xi$, the integral converges absolutely, so the pairing is finite-valued on all of $\Omega^p(M) \times \Omega^p(M)$ — no compact support hypothesis on $\eta, \xi$ is needed. The induced norm is $\|\eta\|_g^2 := \langle\langle \eta, \eta \rangle\rangle_g$. The exterior derivative $d : \Omega^p(M) \to \Omega^{p+1}(M)$ and codifferential $\delta : \Omega^p(M) \to \Omega^{p-1}(M)$ are first-order differential operators on $M$, and the Hodge Laplacian on $p$-forms is \begin{align*} \Delta : \Omega^p(M) &\to \Omega^p(M), \\ \beta &\mapsto d\delta\beta + \delta d\beta. \end{align*}[/step]

[guided]Let $(M, g)$ be a compact oriented Riemannian manifold of dimension $n$ without boundary, and write $\omega_g$ for the Riemannian volume form determined by $g$ and the orientation. The first piece of structure we need is a way to pair two smooth $p$-forms into a single number. Pointwise, the metric $g$ induces a fibrewise inner product $(\cdot, \cdot)_g$ on each fibre of $\Lambda^p T^*M$. To turn this into a global pairing we must integrate over $M$, so we define the global $L^2$ pairing \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*} A natural worry at this stage is whether the integrand is integrable. For arbitrary $\eta, \xi \in \Omega^p(M)$ the function $(\eta, \xi)_g : M \to \mathbb{R}$ is continuous (it is a smooth pairing of smooth sections), and here is where compactness enters for the first time: a continuous function on the compact space $M$ is bounded, so $\int_M (\eta, \xi)_g \, \omega_g$ converges absolutely. Hence the pairing is finite-valued on **all** of $\Omega^p(M) \times \Omega^p(M)$, with no need to restrict to compactly supported forms. This matters because the form $\alpha$ in the theorem is just smooth — it carries no support hypothesis — and we want to apply the pairing to $\alpha$ itself. The induced norm is \begin{align*} \|\eta\|_g^2 := \langle\langle \eta, \eta \rangle\rangle_g. \end{align*} Next we record the differential operators that will appear. The exterior derivative $d : \Omega^p(M) \to \Omega^{p+1}(M)$ raises form degree by one, and the codifferential $\delta : \Omega^p(M) \to \Omega^{p-1}(M)$ lowers it by one. The Hodge Laplacian on $p$-forms is built from these: \begin{align*} \Delta : \Omega^p(M) &\to \Omega^p(M), \\ \beta &\mapsto d\delta\beta + \delta d\beta. \end{align*} The point of this definition is precisely that $\Delta$ mixes $d$ and $\delta$ symmetrically, which is what will let us extract the two equations $d\alpha = 0$ and $\delta\alpha = 0$ from the single equation $\Delta\alpha = 0$ in later steps. The second role of compactness — making the $L^2$-adjoint identity for $d$ apply to all of $\Omega^p(M)$ rather than only to compactly supported forms — will be used in the integration-by-parts step further below; we record it now so that the reader is not surprised when smoothness alone suffices.[/guided]

[step:Prove $d\alpha = 0$ and $\delta\alpha = 0 \Rightarrow \Delta\alpha = 0$ by direct substitution] Suppose $d\alpha = 0$ and $\delta\alpha = 0$. Then \begin{align*} \Delta\alpha = d\delta\alpha + \delta d\alpha = d(0) + \delta(0) = 0, \end{align*} because $d$ and $\delta$ are linear. This direction does not require compactness of $M$ — it is a pointwise algebraic identity valid on any smooth Riemannian manifold. [/step]

[step:Pair $\alpha$ against $\Delta\alpha$ in $L^2$ to convert $\Delta\alpha = 0$ into a sum of squares]Now suppose $\Delta\alpha = 0$. We compute the $L^2$ pairing $\langle\langle \alpha, \Delta\alpha \rangle\rangle_g$ in two different ways and equate the results. On one hand, $\Delta\alpha = 0$ implies \begin{align*} \langle\langle \alpha, \Delta\alpha \rangle\rangle_g = \langle\langle \alpha, 0 \rangle\rangle_g = 0. \end{align*} On the other hand, expanding $\Delta\alpha = d\delta\alpha + \delta d\alpha$ and using bilinearity of the pairing, \begin{align*} \langle\langle \alpha, \Delta\alpha \rangle\rangle_g = \langle\langle \alpha, d\delta\alpha \rangle\rangle_g + \langle\langle \alpha, \delta d\alpha \rangle\rangle_g. \end{align*} We treat each summand separately.[/step]

[guided]The strategy of this step is the standard one for proving "harmonic implies closed and co-closed" on a compact manifold: pair $\alpha$ against $\Delta\alpha$ in $L^2$ and compute the result two ways. One way uses the hypothesis $\Delta\alpha = 0$ directly; the other expands $\Delta = d\delta + \delta d$ and prepares each piece for integration by parts. Equating the two expressions in later steps will collapse the equation into a sum of squared $L^2$-norms. Why is this the right move? The equation $\Delta\alpha = 0$ is a single second-order PDE on $M$, and we want to extract from it **two** first-order equations $d\alpha = 0$ and $\delta\alpha = 0$. A purely pointwise approach cannot achieve this, because $\Delta = d\delta + \delta d$ involves derivatives of derivatives — there is no algebraic identity that splits $\Delta\alpha = 0$ into the two halves. The standard trick is to pass to the global $L^2$ pairing, which converts the second-order operator $\Delta$ into a first-order quadratic form via integration by parts. We compute $\langle\langle \alpha, \Delta\alpha \rangle\rangle_g$ in two ways. **First way: use the hypothesis.** Since $\Delta\alpha = 0$, the pairing of $\alpha$ with the zero $p$-form vanishes by linearity in the second slot: \begin{align*} \langle\langle \alpha, \Delta\alpha \rangle\rangle_g = \langle\langle \alpha, 0 \rangle\rangle_g = 0. \end{align*} **Second way: expand the definition of $\Delta$.** Substituting $\Delta\alpha = d\delta\alpha + \delta d\alpha$ and using bilinearity (specifically, linearity of the pairing in its second slot, applied to the sum $d\delta\alpha + \delta d\alpha$): \begin{align*} \langle\langle \alpha, \Delta\alpha \rangle\rangle_g = \langle\langle \alpha, d\delta\alpha + \delta d\alpha \rangle\rangle_g = \langle\langle \alpha, d\delta\alpha \rangle\rangle_g + \langle\langle \alpha, \delta d\alpha \rangle\rangle_g. \end{align*} We have decomposed the pairing into two summands, each of which contains a composition of $d$ and $\delta$ that is asymmetric — one operator sits on $\alpha$ and the other on the right-hand side. The next step will use the formal adjoint identity for $d$ to flip each pair and produce $\|\delta\alpha\|_g^2$ and $\|d\alpha\|_g^2$ respectively. Equating the two ways of computing the pairing gives $\|\delta\alpha\|_g^2 + \|d\alpha\|_g^2 = 0$, from which non-negativity forces both norms — and hence both forms — to vanish. This is the integration-by-parts mechanism, and the compactness of $M$ without boundary is exactly what kills the boundary terms that would otherwise obstruct the adjoint flip.[/guided]

[step:Apply the adjoint identity for $d$ to each summand]By the [Co-differential is Formal Adjoint of $d$](/theorems/2742), for every $\eta \in \Omega^{p-1}(M)$ and $\xi \in \Omega^p(M)$, \begin{align*} \langle\langle d\eta, \xi \rangle\rangle_g = \langle\langle \eta, \delta\xi \rangle\rangle_g. \tag{A} \end{align*} The hypotheses of theorem 2742 require either that $M$ has no boundary or that the forms be compactly supported. Here $M$ is compact without boundary, so smooth forms on $M$ are automatically supported on the compact set $M$ itself, and (A) applies to all of $\Omega^{p-1}(M)$ and $\Omega^p(M)$ without further restriction. **First summand.** We have $d\delta\alpha = d(\delta\alpha)$ where $\delta\alpha \in \Omega^{p-1}(M)$. By symmetry of the real $L^2$ pairing and (A) applied with $\eta := \delta\alpha$ and $\xi := \alpha$, \begin{align*} \langle\langle \alpha, d\delta\alpha \rangle\rangle_g = \langle\langle d\delta\alpha, \alpha \rangle\rangle_g = \langle\langle d(\delta\alpha), \alpha \rangle\rangle_g = \langle\langle \delta\alpha, \delta\alpha \rangle\rangle_g = \|\delta\alpha\|_g^2. \end{align*} **Second summand.** We have $\delta d\alpha = \delta(d\alpha)$ where $d\alpha \in \Omega^{p+1}(M)$. Apply (A) with $\eta := \alpha \in \Omega^p(M)$ and $\xi := d\alpha \in \Omega^{p+1}(M)$, where (A) is read at the index degree $p \to p+1$: \begin{align*} \langle\langle \alpha, \delta d\alpha \rangle\rangle_g = \langle\langle \alpha, \delta(d\alpha) \rangle\rangle_g = \langle\langle d\alpha, d\alpha \rangle\rangle_g = \|d\alpha\|_g^2. \end{align*}[/step]

[guided]The tool we need is the [Co-differential is Formal Adjoint of $d$](/theorems/2742): for every $\eta \in \Omega^{p-1}(M)$ and $\xi \in \Omega^p(M)$, \begin{align*} \langle\langle d\eta, \xi \rangle\rangle_g = \langle\langle \eta, \delta\xi \rangle\rangle_g. \tag{A} \end{align*} Before applying (A) we verify its hypotheses. Theorem 2742 requires either that $M$ has empty boundary or that the forms $\eta, \xi$ have compact support; this restriction comes from the integration by parts in its proof, where boundary terms must vanish. In our setting $M$ is a closed compact manifold (compact without boundary), so the boundary condition is met automatically and we may apply (A) to **any** smooth forms on $M$ — no support restriction is needed. This is the second appearance of the compactness hypothesis we flagged in Step 1. We now use (A) to evaluate each of the two summands $\langle\langle \alpha, d\delta\alpha \rangle\rangle_g$ and $\langle\langle \alpha, \delta d\alpha \rangle\rangle_g$ produced in the previous step. **First summand: $\langle\langle \alpha, d\delta\alpha \rangle\rangle_g$.** The form inside is $d\delta\alpha = d(\delta\alpha)$ where $\delta\alpha \in \Omega^{p-1}(M)$, so identity (A) applies at the index $p-1 \to p$ with the choice $\eta := \delta\alpha$ and $\xi := \alpha$: \begin{align*} \langle\langle d(\delta\alpha), \alpha \rangle\rangle_g = \langle\langle \delta\alpha, \delta\alpha \rangle\rangle_g. \end{align*} Using symmetry of the real $L^2$ pairing $\langle\langle u, v \rangle\rangle_g = \langle\langle v, u \rangle\rangle_g$ to move $d\delta\alpha$ from the left slot to the right, we find \begin{align*} \langle\langle \alpha, d\delta\alpha \rangle\rangle_g = \langle\langle d\delta\alpha, \alpha \rangle\rangle_g = \langle\langle d(\delta\alpha), \alpha \rangle\rangle_g = \langle\langle \delta\alpha, \delta\alpha \rangle\rangle_g = \|\delta\alpha\|_g^2. \end{align*} The point: applying (A) to a $d\delta$ pair turns the $d$ acting on $\delta\alpha$ on one side into a $\delta$ acting on $\alpha$ on the other, and the two $\delta\alpha$'s collapse into the squared norm. **Second summand: $\langle\langle \alpha, \delta d\alpha \rangle\rangle_g$.** Here the inner form is $\delta d\alpha = \delta(d\alpha)$ where $d\alpha \in \Omega^{p+1}(M)$, so we read (A) at the shifted index $p \to p+1$, with the choice $\eta := \alpha \in \Omega^p(M)$ and $\xi := d\alpha \in \Omega^{p+1}(M)$: \begin{align*} \langle\langle d\alpha, d\alpha \rangle\rangle_g = \langle\langle \alpha, \delta(d\alpha) \rangle\rangle_g. \end{align*} Reading this from right to left, \begin{align*} \langle\langle \alpha, \delta d\alpha \rangle\rangle_g = \langle\langle \alpha, \delta(d\alpha) \rangle\rangle_g = \langle\langle d\alpha, d\alpha \rangle\rangle_g = \|d\alpha\|_g^2. \end{align*} Here the $\delta d$ pair flips into a $d \cdot d$ pair, and the two copies of $d\alpha$ collapse into a squared norm. The two flips are mirror images: the first uses (A) with the $d$ on the left, the second with the $d$ on the right, and in both cases the result is a squared $L^2$-norm of one of the first-order quantities $d\alpha$ or $\delta\alpha$ that we want to show vanish.[/guided]

[step:Combine the two summands to get $\|d\alpha\|_g^2 + \|\delta\alpha\|_g^2 = 0$ and conclude] Combining the previous step with $\langle\langle \alpha, \Delta\alpha \rangle\rangle_g = 0$, \begin{align*} 0 = \langle\langle \alpha, \Delta\alpha \rangle\rangle_g = \langle\langle \alpha, d\delta\alpha \rangle\rangle_g + \langle\langle \alpha, \delta d\alpha \rangle\rangle_g = \|\delta\alpha\|_g^2 + \|d\alpha\|_g^2. \end{align*} Since both $\|\delta\alpha\|_g^2 \ge 0$ and $\|d\alpha\|_g^2 \ge 0$, the vanishing of their sum forces \begin{align*} \|d\alpha\|_g^2 = 0 \quad \text{and} \quad \|\delta\alpha\|_g^2 = 0. \end{align*} The norm $\|\cdot\|_g$ on smooth forms is positive-definite: $\|\eta\|_g^2 = \int_M (\eta, \eta)_g \, \omega_g = 0$ together with $(\eta, \eta)_g \ge 0$ pointwise and continuity of $(\eta, \eta)_g$ on the compact manifold $M$ forces $(\eta, \eta)_g \equiv 0$ on $M$, hence $\eta \equiv 0$. Applying this to $d\alpha$ and $\delta\alpha$, \begin{align*} d\alpha = 0 \quad \text{and} \quad \delta\alpha = 0. \end{align*} This completes both directions of the equivalence: $\Delta\alpha = 0$ if and only if $d\alpha = 0$ and $\delta\alpha = 0$. [/step]