[proofplan] The argument is a single integration-by-parts: writing the global $L^2$ inner product on forms via the pointwise identity $\alpha \wedge \star \beta = (\alpha, \beta)_g \, \omega_g$ on $\Lambda^p$, we use the Leibniz rule $d(\alpha \wedge \star\beta) = d\alpha \wedge \star\beta + (-1)^{p-1} \alpha \wedge d(\star\beta)$ and Stokes' theorem on the closed (or supported-compactly-in-$M$) manifold $M$ to convert $d\alpha \wedge \star\beta$ into $-(-1)^{p-1} \alpha \wedge d(\star\beta)$. The codifferential $\delta$ is then identified by rewriting $d(\star\beta)$ via $\star\star$: applying $\star$ once more and using the [Double Hodge Star](/theorems/2740) sign $(-1)^{p(n-p)}$, we recover the standard expression $\delta = (-1)^{n(p+1)+1} \star d \star$ on $p$-forms (the same definition used to introduce $\delta$ on the page). Pairing the resulting $\alpha \wedge \star \delta\beta$ against the volume form gives $\langle\langle \alpha, \delta\beta \rangle\rangle_g$. [/proofplan]

[step:Set up the $L^2$ pairing on forms via the Hodge star] Let $(M, g)$ be an oriented Riemannian $n$-manifold (assumed without boundary, or with the convention that all integrals below are over compactly supported integrands inside the interior of $M$). Let $\omega_g \in \Omega^n(M)$ denote the Riemannian volume form, and recall that for each degree $k$ the metric $g$ induces a fibrewise inner product $(\cdot, \cdot)_g$ on $\Lambda^k T_x^* M$, with associated $L^2$ pairing \begin{align*} \langle\langle \cdot, \cdot \rangle\rangle_g : \Omega^k_c(M) \times \Omega^k_c(M) &\to \mathbb{R}, \\ (\alpha, \beta) &\mapsto \int_M (\alpha, \beta)_g \, \omega_g, \end{align*} where $\Omega^k_c(M) \subseteq \Omega^k(M)$ denotes compactly supported smooth $k$-forms. Throughout this proof, "compactly supported" is the only integrability assumption — all integrands below are smooth and compactly supported in $M$, so all integrals converge absolutely. The defining identity of the Hodge star \begin{align*} \star : \Lambda^k T_x^* M &\to \Lambda^{n-k} T_x^* M \end{align*} is, pointwise on $M$: \begin{align*} \alpha \wedge \star \beta = (\alpha, \beta)_g \cdot \omega_g \qquad \text{for all } \alpha, \beta \in \Lambda^k T_x^* M. \end{align*} Applied to $\alpha, \beta \in \Omega^k(M)$ and integrated: \begin{align*} \langle\langle \alpha, \beta \rangle\rangle_g = \int_M (\alpha, \beta)_g \, \omega_g = \int_M \alpha \wedge \star \beta. \end{align*} [/step]

[step:Recall the definition of $\delta$ on $p$-forms via $\star$ and $d$] The codifferential $\delta : \Omega^p(M) \to \Omega^{p-1}(M)$ is defined by \begin{align*} \delta \beta = (-1)^{n(p+1) + 1} \star d \star \beta, \end{align*} where $\star$ acts on $\beta$ via $\star : \Omega^p(M) \to \Omega^{n-p}(M)$, then $d$ raises the degree to $n - p + 1$, and the final $\star$ returns to degree $n - (n - p + 1) = p - 1$. The sign $(-1)^{n(p+1)+1}$ is the unique scalar making $\delta$ the formal adjoint of $d$ on $p$-forms; the verification of this sign is precisely what the proof below carries out. By the [Double Hodge Star](/theorems/2740) formula, $\star\star = (-1)^{k(n-k)} \cdot \mathrm{id}$ on $\Lambda^k T^* M$. We will use this twice: once on degree $p$ and once on degree $n - p + 1$. [/step]

[step:Apply Leibniz to $d(\alpha \wedge \star\beta)$ and integrate via Stokes]Let $\alpha \in \Omega^{p-1}_c(M)$ and $\beta \in \Omega^p_c(M)$ (it suffices that $\alpha \wedge \star\beta$ is compactly supported, which holds when either factor is). The exterior derivative satisfies the graded Leibniz rule \begin{align*} d(\alpha \wedge \star\beta) = d\alpha \wedge \star\beta + (-1)^{|\alpha|} \alpha \wedge d(\star\beta) = d\alpha \wedge \star\beta + (-1)^{p-1} \alpha \wedge d(\star\beta), \end{align*} where $|\alpha| = p - 1$. Both terms have degree $n$, so $d(\alpha \wedge \star\beta) \in \Omega^n_c(M)$. Since $\alpha \wedge \star\beta \in \Omega^{n-1}_c(M)$ is compactly supported and $M$ has empty boundary (or equivalently the support is contained in the interior), Stokes' theorem gives \begin{align*} \int_M d(\alpha \wedge \star\beta) = \int_{\partial M} \alpha \wedge \star\beta = 0. \end{align*} Combining with the Leibniz rule and rearranging: \begin{align*} 0 = \int_M d\alpha \wedge \star\beta + (-1)^{p-1} \int_M \alpha \wedge d(\star\beta), \end{align*} hence \begin{align*} \int_M d\alpha \wedge \star\beta = -(-1)^{p-1} \int_M \alpha \wedge d(\star\beta) = (-1)^p \int_M \alpha \wedge d(\star\beta). \end{align*}[/step]

[guided]We want to relate $\int_M d\alpha \wedge \star\beta$ — the integral that will eventually become $\langle\langle d\alpha, \beta \rangle\rangle_g$ — to an integral involving $\alpha$ rather than $d\alpha$. The standard tool for moving a derivative off one factor and onto the other is integration by parts, and on a manifold without boundary, integration by parts is precisely Stokes' theorem applied to the Leibniz expansion of an exact form. We carry out this argument carefully. Fix $\alpha \in \Omega^{p-1}_c(M)$ and $\beta \in \Omega^p_c(M)$. The wedge product $\alpha \wedge \star\beta$ then lies in $\Omega^{n-1}_c(M)$, since $\star\beta \in \Omega^{n-p}(M)$ and $(p-1) + (n-p) = n - 1$. Compact support of either factor forces compact support of the product. Now $d : \Omega^*(M) \to \Omega^{*+1}(M)$ is a graded derivation of degree $+1$, meaning it satisfies the Leibniz rule \begin{align*} d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{|\alpha|} \alpha \wedge d\beta \end{align*} for any homogeneous forms. Specialising with $|\alpha| = p - 1$ and $\beta$ replaced by $\star\beta$: \begin{align*} d(\alpha \wedge \star\beta) = d\alpha \wedge \star\beta + (-1)^{|\alpha|} \alpha \wedge d(\star\beta) = d\alpha \wedge \star\beta + (-1)^{p-1} \alpha \wedge d(\star\beta). \end{align*} Both terms on the right have degree $n$ (the maximal degree on $M$), and both are compactly supported, so $d(\alpha \wedge \star\beta) \in \Omega^n_c(M)$ and is integrable. Now we want to integrate $d(\alpha \wedge \star\beta)$ over $M$ — the answer should be zero, and this is the integration-by-parts step. Why zero? By Stokes' theorem, for a compactly supported $(n-1)$-form $\eta$ on an oriented $n$-manifold $M$, \begin{align*} \int_M d\eta = \int_{\partial M} \eta. \end{align*} We must check the hypotheses. Stokes' theorem requires (i) $M$ oriented (given), (ii) $\eta$ smooth and compactly supported (the form $\alpha \wedge \star\beta$ is smooth because $\alpha, \beta, \star$ all are, and compactly supported because $\alpha$ is). It remains to evaluate the boundary integral. Two scenarios make this vanish: either $M$ has empty boundary (a closed manifold), in which case $\partial M = \emptyset$ and the boundary integral is trivially zero; or $\alpha$ has support strictly inside the interior of $M$, in which case $\alpha \wedge \star\beta$ vanishes on $\partial M$ and the boundary integral is again zero. The hypothesis of the theorem ("$M$ closed, or $\alpha$ compactly supported in the interior") is exactly what we need. Thus \begin{align*} \int_M d(\alpha \wedge \star\beta) = \int_{\partial M} \alpha \wedge \star\beta = 0. \end{align*} Why does this hypothesis matter? If $M$ has non-empty boundary and $\alpha \wedge \star\beta$ does not vanish there, the boundary integral contributes a defect, and the resulting identity $\langle\langle d\alpha, \beta \rangle\rangle_g = \langle\langle \alpha, \delta\beta \rangle\rangle_g$ holds only modulo a boundary term. The clean adjoint identity is a closed-manifold (or compactly-supported) phenomenon. Combining the vanishing with the Leibniz expansion: \begin{align*} 0 = \int_M d(\alpha \wedge \star\beta) = \int_M d\alpha \wedge \star\beta + (-1)^{p-1} \int_M \alpha \wedge d(\star\beta). \end{align*} Solving for the first integral: \begin{align*} \int_M d\alpha \wedge \star\beta = -(-1)^{p-1} \int_M \alpha \wedge d(\star\beta) = (-1)^p \int_M \alpha \wedge d(\star\beta), \end{align*} using $-(-1)^{p-1} = (-1)^p$. The derivative has been transferred from $\alpha$ to $\star\beta$, at the cost of a sign $(-1)^p$. The next step rewrites $d(\star\beta)$ in terms of $\delta\beta$ to expose the adjoint structure.[/guided]

[step:Rewrite $d(\star\beta)$ in terms of $\star \delta\beta$ via the double-star formula]The form $d(\star\beta)$ has degree $n - p + 1$. Apply $\star\star$ to $d(\star\beta)$, which yields back the same form up to a sign by the [Double Hodge Star](/theorems/2740) formula at degree $k = n - p + 1$: \begin{align*} \star\star \big( d (\star \beta) \big) = (-1)^{(n-p+1)(n - (n-p+1))} \cdot d(\star \beta) = (-1)^{(n-p+1)(p-1)} \cdot d(\star \beta). \end{align*} Equivalently, \begin{align*} d(\star \beta) = (-1)^{(n-p+1)(p-1)} \cdot \star \big( \star d \star \beta \big). \end{align*} By the definition of $\delta$ in Step 2, $\star d \star \beta = (-1)^{n(p+1) + 1} \delta \beta$. Substituting, \begin{align*} d(\star \beta) = (-1)^{(n-p+1)(p-1)} \cdot (-1)^{n(p+1) + 1} \cdot \star (\delta \beta). \end{align*} We now check that the combined sign equals $(-1)^p$, which is what the integration-by-parts identity requires. Compute the exponent modulo $2$: \begin{align*} (n - p + 1)(p - 1) + n(p+1) + 1 &= (np - n - p^2 + p + p - 1) + (np + n) + 1 \\ &= np - n - p^2 + 2p - 1 + np + n + 1 \\ &= 2np - p^2 + 2p \\ &\equiv -p^2 + 2p \pmod 2 \\ &\equiv p^2 \pmod 2 \\ &\equiv p \pmod 2. \end{align*} (In the last step we used $p^2 \equiv p \pmod 2$ for any integer $p$.) Therefore \begin{align*} d(\star \beta) = (-1)^p \cdot \star (\delta \beta). \end{align*}[/step]

[guided]At this point we have the identity \begin{align*} \int_M d\alpha \wedge \star\beta = (-1)^p \int_M \alpha \wedge d(\star\beta), \end{align*} but the right-hand side still features $d(\star\beta)$, not $\star(\delta\beta)$. The goal of this step is to massage $d(\star\beta)$ into the form $\star(\delta\beta)$ so that the right-hand side becomes a true $L^2$ pairing. The tool is the double-Hodge-star formula combined with the definition of $\delta$. The form $d(\star\beta)$ has degree $n - p + 1$: indeed, $\star\beta$ has degree $n - p$, and $d$ raises the degree by one. Applying $\star\star$ to a $(n-p+1)$-form returns it up to a sign computed by the [Double Hodge Star](/theorems/2740) formula, which states $\star\star = (-1)^{k(n-k)} \cdot \mathrm{id}$ on $\Lambda^k T^*M$. Specialising at $k = n - p + 1$ (so that $n - k = p - 1$): \begin{align*} \star\star \big( d(\star\beta) \big) = (-1)^{(n-p+1)(p-1)} \cdot d(\star\beta). \end{align*} We can solve this for $d(\star\beta)$ — the sign $(-1)^{(n-p+1)(p-1)}$ is its own inverse modulo signs, so multiplying both sides by it: \begin{align*} d(\star\beta) = (-1)^{(n-p+1)(p-1)} \cdot \star\big(\star d \star \beta\big). \end{align*} Now we recognise $\star d \star \beta$ from the definition of $\delta$ in Step 2. Since $\delta\beta = (-1)^{n(p+1) + 1} \star d \star \beta$, we can solve for $\star d \star \beta$: \begin{align*} \star d \star \beta = (-1)^{n(p+1) + 1} \delta\beta. \end{align*} (The sign $(-1)^{n(p+1)+1}$ squares to $1$, so it appears unchanged when we transpose it across the equation.) Substituting: \begin{align*} d(\star\beta) = (-1)^{(n-p+1)(p-1)} \cdot (-1)^{n(p+1) + 1} \cdot \star(\delta\beta). \end{align*} It remains to verify that the combined sign $(-1)^{(n-p+1)(p-1) + n(p+1) + 1}$ equals $(-1)^p$ — this is the parity check that ensures the integration-by-parts identity will close up cleanly. We compute the exponent modulo $2$ by direct expansion: \begin{align*} (n - p + 1)(p - 1) + n(p+1) + 1 &= (np - n - p^2 + p + p - 1) + (np + n) + 1 \\ &= np - n - p^2 + 2p - 1 + np + n + 1 \\ &= 2np - p^2 + 2p \\ &\equiv -p^2 + 2p \pmod 2 \\ &\equiv p^2 \pmod 2 \\ &\equiv p \pmod 2. \end{align*} The simplifications used at each step: $2np \equiv 0 \pmod 2$ since it is even; $2p \equiv 0 \pmod 2$; $-p^2 \equiv p^2 \pmod 2$ since signs do not change parity; and $p^2 \equiv p \pmod 2$ for any integer (this is a standard observation: $p$ and $p^2$ are both even or both odd, since $p$ even gives $p^2$ even and $p$ odd gives $p^2$ odd). Therefore \begin{align*} d(\star\beta) = (-1)^p \cdot \star(\delta\beta). \end{align*} This is the local identity that makes the adjoint formula work. The sign $(-1)^{n(p+1)+1}$ in the definition of $\delta$ was engineered precisely so that this cancellation produces $(-1)^p$. Different textbooks adopt different conventions for $\delta$, absorbing different combinations of signs from $\star$ and $d$; all yield the same operator and the same adjoint identity, but the bookkeeping differs. We have used the convention stated on the codifferential page in this chapter.[/guided]

[step:Substitute and identify the $L^2$ pairing] Substituting $d(\star\beta) = (-1)^p \cdot \star(\delta\beta)$ into the result of Step 3: \begin{align*} \int_M d\alpha \wedge \star\beta = (-1)^p \int_M \alpha \wedge d(\star\beta) = (-1)^p \cdot (-1)^p \int_M \alpha \wedge \star(\delta\beta) = \int_M \alpha \wedge \star(\delta\beta), \end{align*} using $(-1)^p \cdot (-1)^p = (-1)^{2p} = 1$. By the $L^2$ pairing identity from Step 1, applied at degrees $p$ and $p - 1$ respectively, \begin{align*} \int_M d\alpha \wedge \star\beta = \langle\langle d\alpha, \beta \rangle\rangle_g, \qquad \int_M \alpha \wedge \star(\delta\beta) = \langle\langle \alpha, \delta\beta \rangle\rangle_g. \end{align*} (The first uses $d\alpha \in \Omega^p_c(M)$ paired with $\beta \in \Omega^p_c(M)$; the second uses $\alpha \in \Omega^{p-1}_c(M)$ paired with $\delta\beta \in \Omega^{p-1}(M)$, with the integrand $\alpha \wedge \star(\delta\beta)$ compactly supported because $\alpha$ is.) Combining: \begin{align*} \langle\langle d\alpha, \beta \rangle\rangle_g = \langle\langle \alpha, \delta\beta \rangle\rangle_g. \end{align*} This is the claimed adjoint identity, and shows that $\delta$ is the formal $L^2$ adjoint of $d$. The proof is complete. [/step]