[proofplan] The first refined decomposition $\Omega^p(M) = \mathcal{H}^p \oplus d\delta\Omega^p(M) \oplus \delta d\Omega^p(M)$ refines the decomposition from the [Hodge Decomposition](/theorems/2745) by splitting $\Delta\Omega^p(M) = d\delta\Omega^p(M) + \delta d\Omega^p(M)$ into a sum of two mutually orthogonal pieces. The orthogonality $d\delta\Omega^p(M) \perp \delta d\Omega^p(M)$ comes from $d^2 = 0$ via two adjoint flips. The second refined decomposition $\Omega^p(M) = \mathcal{H}^p \oplus d\Omega^{p-1}(M) \oplus \delta\Omega^{p+1}(M)$ then follows from the first by enlarging the second and third summands ($d\delta\Omega^p(M) \subseteq d\Omega^{p-1}(M)$ and $\delta d\Omega^p(M) \subseteq \delta\Omega^{p+1}(M)$) and showing that the inclusions are equalities up to elements already in $\mathcal{H}^p$, using that any closed (resp. co-closed) form decomposes via the Hodge decomposition with vanishing $\delta d$ (resp. $d\delta$) component. [/proofplan]

[step:Recall the Hodge decomposition and set up the global $L^2$ inner product] Let $(M, g)$ be a compact oriented Riemannian manifold without boundary, $\omega_g$ 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, with associated norm $\|\eta\|_g$. By the [Hodge Decomposition](/theorems/2745), \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus \Delta\Omega^p(M), \tag{H} \end{align*} as an orthogonal direct sum. We will use the formal adjoint identity \begin{align*} \langle\langle d\eta, \xi \rangle\rangle_g = \langle\langle \eta, \delta\xi \rangle\rangle_g \tag{A} \end{align*} from the [Co-differential is Formal Adjoint of $d$](/theorems/2742), which holds for all $\eta \in \Omega^{p-1}(M)$ and $\xi \in \Omega^p(M)$ on the closed manifold $M$ (no compact support hypothesis is needed because $M$ itself is compact). [/step]

[step:Show $d\delta\Omega^p(M)$ and $\delta d\Omega^p(M)$ are mutually $L^2$-orthogonal]Let $\alpha, \beta \in \Omega^p(M)$. We compute \begin{align*} \langle\langle d\delta\alpha, \delta d\beta \rangle\rangle_g. \end{align*} By symmetry of the $L^2$ pairing, \begin{align*} \langle\langle d\delta\alpha, \delta d\beta \rangle\rangle_g = \langle\langle \delta d\beta, d\delta\alpha \rangle\rangle_g. \end{align*} Setting $\eta := \tau \in \Omega^p(M)$ and $\xi := \sigma \in \Omega^{p+1}(M)$ in (A) and using symmetry of the pairing yields the complementary identity \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*} Applying (A') with $\sigma := d\beta \in \Omega^{p+1}(M)$ and $\tau := d\delta\alpha \in \Omega^p(M)$, \begin{align*} \langle\langle \delta d\beta, d\delta\alpha \rangle\rangle_g = \langle\langle d\beta, d(d\delta\alpha) \rangle\rangle_g = \langle\langle d\beta, d^2\delta\alpha \rangle\rangle_g = \langle\langle d\beta, 0 \rangle\rangle_g = 0, \end{align*} where we used $d^2 = 0$. Therefore \begin{align*} \langle\langle d\delta\alpha, \delta d\beta \rangle\rangle_g = 0 \tag{$\perp_1$} \end{align*} for all $\alpha, \beta \in \Omega^p(M)$. The two subspaces $d\delta\Omega^p(M)$ and $\delta d\Omega^p(M)$ are therefore mutually orthogonal in $\Omega^p(M)$.[/step]

[guided]We want to show that the two summands $d\delta\Omega^p(M)$ and $\delta d\Omega^p(M)$ inside the Laplacian image are mutually $L^2$-orthogonal. Why should this be true at all? The Hodge Laplacian factors as $\Delta = d\delta + \delta d$, and the only structural identity available is $d^2 = 0$ (and its adjoint $\delta^2 = 0$). The plan is to use the formal-adjoint relation (A) to push $\delta$ across the $L^2$ pairing until a $d^2$ collides with itself, killing the inner product. Let $\alpha, \beta \in \Omega^p(M)$ be arbitrary, and consider the pairing \begin{align*} \langle\langle d\delta\alpha, \delta d\beta \rangle\rangle_g. \end{align*} The $L^2$ pairing on real-valued forms is symmetric, so we may rewrite this as \begin{align*} \langle\langle d\delta\alpha, \delta d\beta \rangle\rangle_g = \langle\langle \delta d\beta, d\delta\alpha \rangle\rangle_g. \end{align*} Now we need to move $\delta$ off of $d\beta$. The formal-adjoint identity (A) reads $\langle\langle d\eta, \xi \rangle\rangle_g = \langle\langle \eta, \delta\xi \rangle\rangle_g$. Reading it from right to left in the symmetric variable convention, and substituting $\eta := \tau \in \Omega^p(M)$ and $\xi := \sigma \in \Omega^{p+1}(M)$, we obtain the complementary statement \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*} This is the same theorem as (A), just relabelled — the hypotheses (compact closed manifold) are inherited from (A) and require no separate verification. We now apply (A') with $\sigma := d\beta \in \Omega^{p+1}(M)$ and $\tau := d\delta\alpha \in \Omega^p(M)$, which are valid choices because $d : \Omega^p(M) \to \Omega^{p+1}(M)$ and $d\delta : \Omega^p(M) \to \Omega^p(M)$: \begin{align*} \langle\langle \delta d\beta, d\delta\alpha \rangle\rangle_g = \langle\langle d\beta, d(d\delta\alpha) \rangle\rangle_g = \langle\langle d\beta, d^2\delta\alpha \rangle\rangle_g = \langle\langle d\beta, 0 \rangle\rangle_g = 0, \end{align*} where the third equality is $d \circ d = 0$. The cancellation that does the work is $d^2 = 0$ — everything else is bookkeeping that converts this pointwise identity into an $L^2$ statement. Therefore \begin{align*} \langle\langle d\delta\alpha, \delta d\beta \rangle\rangle_g = 0 \tag{$\perp_1$} \end{align*} for all $\alpha, \beta \in \Omega^p(M)$, which says exactly that $d\delta\Omega^p(M) \perp \delta d\Omega^p(M)$ in the $L^2$ inner product. This is the abstract reason why the Laplacian image splits orthogonally into two pieces: each element of $\delta d\Omega^p$ pulls back across the adjoint to a $d^2$-image, which vanishes.[/guided]

[step:Show $d\delta\Omega^p(M) + \delta d\Omega^p(M) = \Delta\Omega^p(M)$ and conclude the first refined decomposition]By definition, \begin{align*} \Delta\gamma = d\delta\gamma + \delta d\gamma \in d\delta\Omega^p(M) + \delta d\Omega^p(M) \end{align*} for every $\gamma \in \Omega^p(M)$, so $\Delta\Omega^p(M) \subseteq d\delta\Omega^p(M) + \delta d\Omega^p(M)$. Conversely, let $\eta = d\delta\alpha + \delta d\beta$ for some $\alpha, \beta \in \Omega^p(M)$. We do not yet know that $\eta \in \Delta\Omega^p(M)$ since the two arguments $\alpha$ and $\beta$ are independent. However, the orthogonal decomposition (H) applied to $\alpha$ and $\beta$ separately gives \begin{align*} \alpha &= \alpha_\mathcal{H} + \Delta\alpha', \\ \beta &= \beta_\mathcal{H} + \Delta\beta', \end{align*} with $\alpha_\mathcal{H}, \beta_\mathcal{H} \in \mathcal{H}^p$ and $\alpha', \beta' \in \Omega^p(M)$. Since $\alpha_\mathcal{H}, \beta_\mathcal{H}$ are harmonic, the [Harmonic Iff Closed and Co-closed](/theorems/2744) gives $d\alpha_\mathcal{H} = \delta\alpha_\mathcal{H} = d\beta_\mathcal{H} = \delta\beta_\mathcal{H} = 0$. Thus \begin{align*} d\delta\alpha = d\delta\Delta\alpha' \quad \text{and} \quad \delta d\beta = \delta d\Delta\beta'. \end{align*} Now the operators $d\delta$ and $\delta d$ each commute with $\Delta = d\delta + \delta d$ as operators on $\Omega^p(M)$, since $d^2 = 0$ and $\delta^2 = 0$ make $d\delta \cdot \delta d = 0 = \delta d \cdot d\delta$, so \begin{align*} (d\delta)\Delta &= (d\delta)(d\delta + \delta d) = d\delta d\delta + d\delta\delta d = d\delta d\delta = (d\delta + \delta d)(d\delta) = \Delta(d\delta), \\ (\delta d)\Delta &= \Delta(\delta d) \qquad \text{by the same calculation.} \end{align*} Therefore $d\delta\Delta\alpha' = \Delta(d\delta\alpha')$ and $\delta d\Delta\beta' = \Delta(\delta d\beta')$. Hence \begin{align*} \eta = d\delta\alpha + \delta d\beta = \Delta(d\delta\alpha') + \Delta(\delta d\beta') = \Delta(d\delta\alpha' + \delta d\beta') \in \Delta\Omega^p(M). \end{align*} Combining, \begin{align*} d\delta\Omega^p(M) + \delta d\Omega^p(M) = \Delta\Omega^p(M). \end{align*} Together with the orthogonality $(\perp_1)$ and the orthogonal direct sum (H), we obtain \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus d\delta\Omega^p(M) \oplus \delta d\Omega^p(M), \tag{D1} \end{align*} as an orthogonal direct sum.[/step]

[guided]We need to show that the two-argument sum $d\delta\Omega^p(M) + \delta d\Omega^p(M)$ coincides with the single-argument image $\Delta\Omega^p(M)$, and then assemble the orthogonal direct sum decomposition. *Forward inclusion $\Delta\Omega^p(M) \subseteq d\delta\Omega^p(M) + \delta d\Omega^p(M)$.* This direction is immediate from the very definition of $\Delta$. For any $\gamma \in \Omega^p(M)$, \begin{align*} \Delta\gamma = d\delta\gamma + \delta d\gamma \in d\delta\Omega^p(M) + \delta d\Omega^p(M). \end{align*} *Reverse inclusion $d\delta\Omega^p(M) + \delta d\Omega^p(M) \subseteq \Delta\Omega^p(M)$.* This is the non-obvious direction. The asymmetry is that elements of $\Delta\Omega^p$ have a single seed $\gamma$, whereas elements of the sum have two independent seeds $\alpha, \beta$. We must collapse the two seeds into one. Let $\eta = d\delta\alpha + \delta d\beta$ for some $\alpha, \beta \in \Omega^p(M)$. The strategy is: first kill the harmonic components of $\alpha$ and $\beta$ (which contribute zero to $d\delta\alpha$ and $\delta d\beta$ respectively), so that we may assume $\alpha, \beta \in \Delta\Omega^p$; then commute $d\delta$ and $\delta d$ across $\Delta$ to extract a single $\Delta$. Apply the orthogonal decomposition (H), $\Omega^p(M) = \mathcal{H}^p \oplus \Delta\Omega^p(M)$, separately to $\alpha$ and to $\beta$ (the hypotheses for (H) — $M$ compact oriented Riemannian without boundary — are exactly those in scope): \begin{align*} \alpha &= \alpha_\mathcal{H} + \Delta\alpha', \\ \beta &= \beta_\mathcal{H} + \Delta\beta', \end{align*} with $\alpha_\mathcal{H}, \beta_\mathcal{H} \in \mathcal{H}^p$ and $\alpha', \beta' \in \Omega^p(M)$. By [Harmonic Iff Closed and Co-closed](/theorems/2744), $\alpha_\mathcal{H}$ harmonic implies $d\alpha_\mathcal{H} = 0$ and $\delta\alpha_\mathcal{H} = 0$, and likewise for $\beta_\mathcal{H}$. Why does this help? Because $d\delta$ applied to $\alpha_\mathcal{H}$ vanishes — it kills $\delta\alpha_\mathcal{H}$ in the inner step — and analogously for $\delta d\beta_\mathcal{H}$. Therefore \begin{align*} d\delta\alpha = d\delta\Delta\alpha' \quad \text{and} \quad \delta d\beta = \delta d\Delta\beta'. \end{align*} Now we want to commute the outer operator past $\Delta$. The crucial calculation is that $d\delta$ and $\delta d$ each commute with $\Delta = d\delta + \delta d$, because the cross terms vanish: $d\delta \cdot \delta d$ contains $\delta\delta = \delta^2 = 0$, and $\delta d \cdot d\delta$ contains $dd = d^2 = 0$. Explicitly, \begin{align*} (d\delta)\Delta &= (d\delta)(d\delta + \delta d) = d\delta d\delta + d\delta\delta d = d\delta d\delta = (d\delta + \delta d)(d\delta) = \Delta(d\delta), \\ (\delta d)\Delta &= \Delta(\delta d) \qquad \text{by the symmetric calculation using $d^2 = 0$.} \end{align*} This commutation is what allows us to pull $\Delta$ outside: $d\delta\Delta\alpha' = \Delta(d\delta\alpha')$ and $\delta d\Delta\beta' = \Delta(\delta d\beta')$. Substituting back, \begin{align*} \eta = d\delta\alpha + \delta d\beta = \Delta(d\delta\alpha') + \Delta(\delta d\beta') = \Delta(d\delta\alpha' + \delta d\beta') \in \Delta\Omega^p(M). \end{align*} The two seeds $\alpha', \beta'$ have been merged into the single seed $d\delta\alpha' + \delta d\beta'$, completing the reverse inclusion. Combining both inclusions, \begin{align*} d\delta\Omega^p(M) + \delta d\Omega^p(M) = \Delta\Omega^p(M). \end{align*} *Assembling the orthogonal direct sum.* By Step 2, $d\delta\Omega^p(M) \perp \delta d\Omega^p(M)$, so the sum on the left is in fact an orthogonal direct sum $d\delta\Omega^p(M) \oplus \delta d\Omega^p(M)$. Together with (H), we obtain \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus d\delta\Omega^p(M) \oplus \delta d\Omega^p(M), \tag{D1} \end{align*} as an orthogonal direct sum. (The orthogonalities $\mathcal{H}^p \perp d\delta\Omega^p(M)$ and $\mathcal{H}^p \perp \delta d\Omega^p(M)$ follow from $\mathcal{H}^p \perp \Delta\Omega^p(M)$ in (H), since $d\delta\Omega^p(M), \delta d\Omega^p(M) \subseteq \Delta\Omega^p(M)$ by the equality just established.)[/guided]

[step:Show $d\Omega^{p-1}(M) = d\delta\Omega^p(M)$ and $\delta\Omega^{p+1}(M) = \delta d\Omega^p(M)$ via Hodge decomposition at adjacent degrees]We now derive the second refined decomposition. Note first the inclusion \begin{align*} d\delta\Omega^p(M) \subseteq d\Omega^{p-1}(M), \end{align*} since $d\delta\alpha = d(\delta\alpha)$ with $\delta\alpha \in \Omega^{p-1}(M)$. Conversely, let $d\sigma \in d\Omega^{p-1}(M)$ with $\sigma \in \Omega^{p-1}(M)$. Apply the [Hodge Decomposition](/theorems/2745) at degree $p-1$: \begin{align*} \sigma = \sigma_\mathcal{H} + \Delta\tau, \end{align*} with $\sigma_\mathcal{H} \in \mathcal{H}^{p-1}$ and $\tau \in \Omega^{p-1}(M)$. Then $d\sigma_\mathcal{H} = 0$ by [Harmonic Iff Closed and Co-closed](/theorems/2744), so \begin{align*} d\sigma = d\Delta\tau = d(d\delta\tau + \delta d\tau) = dd\delta\tau + d\delta d\tau = d\delta(d\tau), \end{align*} where we used $d^2 = 0$ in the form $d \circ d = 0$ to drop the first term. Thus $d\sigma = d\delta(d\tau) \in d\delta\Omega^p(M)$. Therefore \begin{align*} d\Omega^{p-1}(M) = d\delta\Omega^p(M). \tag{E1} \end{align*} By the analogous argument at degree $p+1$, \begin{align*} \delta\Omega^{p+1}(M) = \delta d\Omega^p(M). \tag{E2} \end{align*} The argument: let $\delta\rho \in \delta\Omega^{p+1}(M)$ with $\rho \in \Omega^{p+1}(M)$. Apply the Hodge decomposition at degree $p+1$: $\rho = \rho_\mathcal{H} + \Delta\mu$ with $\rho_\mathcal{H} \in \mathcal{H}^{p+1}$, $\mu \in \Omega^{p+1}(M)$. Then $\delta\rho_\mathcal{H} = 0$ and \begin{align*} \delta\rho = \delta\Delta\mu = \delta(d\delta\mu + \delta d\mu) = \delta d\delta\mu + \delta\delta d\mu = \delta d(\delta\mu), \end{align*} using $\delta^2 = 0$. So $\delta\rho \in \delta d\Omega^p(M)$.[/step]

[guided]The goal is to identify the two refined summands $d\delta\Omega^p(M)$ and $\delta d\Omega^p(M)$ from (D1) with the more familiar spaces $d\Omega^{p-1}(M)$ and $\delta\Omega^{p+1}(M)$, which are the images of $d$ and $\delta$ from adjacent degrees. The forward inclusions are immediate by inspection; the reverse inclusions require a Hodge decomposition at the adjacent degree. The structural tool that makes the reverse inclusions work is $d^2 = 0$ (for E1) and $\delta^2 = 0$ (for E2): each of these collapses one of the two terms in $\Delta = d\delta + \delta d$, leaving the desired image. *Proof of (E1): $d\Omega^{p-1}(M) = d\delta\Omega^p(M)$.* For the forward inclusion $d\delta\Omega^p(M) \subseteq d\Omega^{p-1}(M)$, let $d\delta\alpha \in d\delta\Omega^p(M)$ for some $\alpha \in \Omega^p(M)$. Since $\delta : \Omega^p(M) \to \Omega^{p-1}(M)$, the form $\delta\alpha$ lies in $\Omega^{p-1}(M)$, so \begin{align*} d\delta\Omega^p(M) \subseteq d\Omega^{p-1}(M). \end{align*} For the reverse inclusion, let $d\sigma \in d\Omega^{p-1}(M)$ with $\sigma \in \Omega^{p-1}(M)$. We need to exhibit some $\tilde\alpha \in \Omega^p(M)$ with $d\sigma = d\delta\tilde\alpha$. The natural move is to decompose $\sigma$ via Hodge at degree $p-1$, so that $\sigma$ is split into a closed part (which $d$ kills) and a Laplacian-image part (where $\Delta = d\delta + \delta d$ provides the structure to factor through $d\delta$). Apply the [Hodge Decomposition](/theorems/2745) at degree $p-1$ — the hypotheses ($M$ compact oriented Riemannian without boundary) are exactly those in scope, so no further verification is needed: \begin{align*} \sigma = \sigma_\mathcal{H} + \Delta\tau, \end{align*} with $\sigma_\mathcal{H} \in \mathcal{H}^{p-1}$ and $\tau \in \Omega^{p-1}(M)$. Since $\sigma_\mathcal{H}$ is harmonic, [Harmonic Iff Closed and Co-closed](/theorems/2744) at degree $p-1$ gives $d\sigma_\mathcal{H} = 0$. Now compute: \begin{align*} d\sigma = d\sigma_\mathcal{H} + d\Delta\tau = 0 + d(d\delta\tau + \delta d\tau) = d^2\delta\tau + d\delta(d\tau) = d\delta(d\tau). \end{align*} The first term $d^2\delta\tau$ vanishes by $d^2 = 0$ — this is the moment where $d^2 = 0$ does the structural work. Setting $\tilde\alpha := d\tau \in \Omega^p(M)$, we have written $d\sigma = d\delta\tilde\alpha \in d\delta\Omega^p(M)$, which gives the reverse inclusion. Combining, \begin{align*} d\Omega^{p-1}(M) = d\delta\Omega^p(M). \tag{E1} \end{align*} *Proof of (E2): $\delta\Omega^{p+1}(M) = \delta d\Omega^p(M)$.* The argument is the formal mirror at degree $p+1$, using $\delta^2 = 0$ in place of $d^2 = 0$. The forward inclusion is immediate: $\delta d\beta = \delta(d\beta)$ with $d\beta \in \Omega^{p+1}(M)$, so $\delta d\Omega^p(M) \subseteq \delta\Omega^{p+1}(M)$. For the reverse inclusion, let $\delta\rho \in \delta\Omega^{p+1}(M)$ with $\rho \in \Omega^{p+1}(M)$. Apply the [Hodge Decomposition](/theorems/2745) at degree $p+1$ (same hypotheses as before): $\rho = \rho_\mathcal{H} + \Delta\mu$ with $\rho_\mathcal{H} \in \mathcal{H}^{p+1}$ and $\mu \in \Omega^{p+1}(M)$. By [Harmonic Iff Closed and Co-closed](/theorems/2744) at degree $p+1$, $\rho_\mathcal{H}$ harmonic implies $\delta\rho_\mathcal{H} = 0$. Therefore \begin{align*} \delta\rho = \delta\rho_\mathcal{H} + \delta\Delta\mu = 0 + \delta(d\delta\mu + \delta d\mu) = \delta d\delta\mu + \delta^2 d\mu = \delta d(\delta\mu), \end{align*} where the last term $\delta^2 d\mu$ vanishes by $\delta^2 = 0$ — the mirror cancellation. Setting $\tilde\beta := \delta\mu \in \Omega^p(M)$, we have $\delta\rho = \delta d\tilde\beta \in \delta d\Omega^p(M)$, so \begin{align*} \delta\Omega^{p+1}(M) = \delta d\Omega^p(M). \tag{E2} \end{align*} The two identifications (E1) and (E2) are exactly what we need to convert (D1) into the second refined decomposition.[/guided]

[step:Combine to obtain the second refined decomposition]Substituting (E1) and (E2) into (D1): \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus d\delta\Omega^p(M) \oplus \delta d\Omega^p(M) = \mathcal{H}^p \oplus d\Omega^{p-1}(M) \oplus \delta\Omega^{p+1}(M), \end{align*} as an orthogonal direct sum. This completes the proof: both refined orthogonal direct sum decompositions hold under the same hypotheses as the [Hodge Decomposition](/theorems/2745).[/step]

[guided]We now have all the pieces to assemble the second refined decomposition. The first decomposition (D1) splits $\Omega^p(M)$ into harmonic, $d\delta$-image, and $\delta d$-image pieces; the identifications (E1) and (E2) say that the latter two coincide with the more familiar images $d\Omega^{p-1}(M)$ and $\delta\Omega^{p+1}(M)$ respectively. Substituting (E1) into the second summand and (E2) into the third summand of (D1): \begin{align*} \Omega^p(M) = \mathcal{H}^p \oplus d\delta\Omega^p(M) \oplus \delta d\Omega^p(M) = \mathcal{H}^p \oplus d\Omega^{p-1}(M) \oplus \delta\Omega^{p+1}(M). \end{align*} Why is this still an *orthogonal* direct sum? Orthogonality is preserved under set-equality of summands: if $V_1 \perp V_2$ and $V_2 = V_2'$, then $V_1 \perp V_2'$. The orthogonalities $\mathcal{H}^p \perp d\delta\Omega^p(M)$, $\mathcal{H}^p \perp \delta d\Omega^p(M)$, and $d\delta\Omega^p(M) \perp \delta d\Omega^p(M)$ from (D1) therefore transfer through (E1) and (E2) to give $\mathcal{H}^p \perp d\Omega^{p-1}(M)$, $\mathcal{H}^p \perp \delta\Omega^{p+1}(M)$, and $d\Omega^{p-1}(M) \perp \delta\Omega^{p+1}(M)$. So the second refined decomposition holds as an orthogonal direct sum under exactly the same hypotheses as the [Hodge Decomposition](/theorems/2745). The strategic content is: $d^2 = 0$ and $\delta^2 = 0$ together with Hodge decomposition at adjacent degrees are precisely what is needed to upgrade the abstract Laplacian-image splitting into the geometrically natural splitting via $d\Omega^{p-1}(M)$ and $\delta\Omega^{p+1}(M)$.[/guided]