[proofplan] The argument has two parts. For (1), we use that the irreducible decomposition $\bigwedge^k T^* M = \bigoplus_i \Lambda_i^k$ is *parallel*: each $\Lambda_i^k$ is preserved by $\nabla$ because the holonomy decomposition into irreducibles is built from $\mathrm{Hol}_x(g)$-invariant subspaces, which by the [Fundamental Principle of Riemannian Holonomy](/theorems/2764) correspond to parallel sub-bundles. Since $d$ and $\delta$ are first-order operators built from the Levi-Civita connection (and the metric, for $\delta$), the Hodge Laplacian $\Delta = d\delta + \delta d$ commutes with the parallel projection onto each $\Lambda_i^k$. So $\Delta$ preserves $\Omega_i^k(M)$. For (2), we apply Hodge decomposition: every cohomology class is represented by a unique harmonic form. The orthogonal projection of a harmonic form onto $\Omega_i^k(M)$ — implemented by the parallel sub-bundle projection — is again harmonic (because $\Delta$ commutes with the projection by part (1)), and the components are mutually orthogonal in $L^2$ since the irreducible factors are pointwise orthogonal. This gives the cohomology splitting. [/proofplan]

[step:Show that each $\Lambda_i^k$ is a parallel sub-bundle of $\bigwedge^k T^*M$]Fix a point $x \in M$ and the decomposition \begin{align*} \bigwedge^k T_x^* M = \bigoplus_i (\Lambda_i^k)_x \end{align*} into irreducible $\mathrm{Hol}_x(g)$-representations. Let $\pi_i^x : \bigwedge^k T_x^* M \to (\Lambda_i^k)_x$ be the corresponding orthogonal projection (orthogonal with respect to the $g$-induced inner product on $\bigwedge^k T_x^* M$). Each $\pi_i^x$ is an element of $\mathrm{End}(\bigwedge^k T_x^* M)$. [claim:Each projection $\pi_i^x$ is invariant under the action of $\mathrm{Hol}_x(g)$] [proof] The decomposition $\bigwedge^k T_x^* M = \bigoplus_i (\Lambda_i^k)_x$ is, by hypothesis, the decomposition into $\mathrm{Hol}_x(g)$-irreducibles, so each summand $(\Lambda_i^k)_x$ is preserved by every $A \in \mathrm{Hol}_x(g)$. Since $\mathrm{Hol}_x(g) \subseteq \mathrm{O}(T_x M, g_x)$ and the action on $\bigwedge^k T_x^* M$ is by orthogonal transformations (induced from an orthogonal action on $T_x M$, see [Parallel Transport, Holonomy and Restricted Holonomy](/theorems/2762)), the orthogonal complements are also preserved. Therefore each orthogonal projection $\pi_i^x$ commutes with every $A \in \mathrm{Hol}_x(g)$: \begin{align*} A \circ \pi_i^x = \pi_i^x \circ A \qquad \text{for all } A \in \mathrm{Hol}_x(g). \end{align*} Equivalently, $A \pi_i^x A^{-1} = \pi_i^x$ for all $A \in \mathrm{Hol}_x(g)$. Viewing $\pi_i^x \in \mathrm{End}(\bigwedge^k T_x^* M) \cong \bigwedge^k T_x^* M \otimes \bigwedge^k T_x M$ as a tensor at $x$, this is exactly Hol-invariance of $\pi_i^x$. [/proof] [/claim] By the [Fundamental Principle of Riemannian Holonomy](/theorems/2764) applied to the tensor $\pi_i^x$ at $x$ (which is Hol-invariant by the claim), there is a parallel section $\pi_i \in \Gamma(\mathrm{End}(\bigwedge^k T^*M))$ with $\pi_i|_x = \pi_i^x$ and $\nabla \pi_i = 0$. By construction (parallel transport from $x$), the value $(\pi_i)_y$ at any other $y \in M$ is again an orthogonal projection, with image \begin{align*} (\Lambda_i^k)_y := \mathrm{im}((\pi_i)_y) \subseteq \bigwedge^k T_y^* M. \end{align*} The collection $\Lambda_i^k = \bigsqcup_y (\Lambda_i^k)_y$ is a smooth sub-bundle of $\bigwedge^k T^* M$ (smoothness follows from smoothness of $\pi_i$ as a section of $\mathrm{End}(\bigwedge^k T^*M)$; see [Fundamental Principle of Riemannian Holonomy](/theorems/2764)). The defining property $\nabla \pi_i = 0$ says exactly that the sub-bundle $\Lambda_i^k$ is **parallel**: for any vector field $X$ and any section $\alpha \in \Gamma(\Lambda_i^k)$, \begin{align*} \nabla_X \alpha = \nabla_X(\pi_i \alpha) = (\nabla_X \pi_i) \alpha + \pi_i(\nabla_X \alpha) = \pi_i(\nabla_X \alpha) \in \Gamma(\Lambda_i^k). \end{align*} So $\Lambda_i^k$ is preserved by $\nabla$, justifying the use of $\Omega_i^k(M) := \Gamma(\Lambda_i^k)$ as a parallel-stable subspace of $\Omega^k(M)$.[/step]

[guided]The first key insight is that the holonomy-irreducible decomposition at *one point* automatically extends to a decomposition into parallel sub-bundles *globally*. The bridge between local (one-point) and global (parallel section) data is the [Fundamental Principle of Riemannian Holonomy](/theorems/2764): a $\mathrm{Hol}_x(g)$-invariant tensor at $x$ corresponds bijectively to a parallel tensor field on $M$. To use it, we need a Hol-invariant tensor — and the natural candidate is the orthogonal projection. Fix $x \in M$. The decomposition into Hol-irreducibles is, by hypothesis, \begin{align*} \bigwedge^k T_x^* M = \bigoplus_i (\Lambda_i^k)_x. \end{align*} Let $\pi_i^x : \bigwedge^k T_x^* M \to (\Lambda_i^k)_x$ be the orthogonal projection (orthogonal with respect to the $g$-induced inner product on $\bigwedge^k T_x^* M$). *Why is $\pi_i^x$ Hol-invariant?* Hol acts on $\bigwedge^k T_x^* M$ orthogonally — this is because $\mathrm{Hol}_x(g) \subseteq \mathrm{O}(T_x M, g_x)$ and the induced action on exterior powers is orthogonal, see [Parallel Transport, Holonomy and Restricted Holonomy](/theorems/2762). Each summand $(\Lambda_i^k)_x$ is, by assumption, Hol-stable. Orthogonal actions stabilise orthogonal complements as well, so the kernel of $\pi_i^x$ (the sum of the other summands) is also Hol-stable. An orthogonal projection onto a Hol-stable subspace with a Hol-stable kernel commutes with the Hol-action: for every $A \in \mathrm{Hol}_x(g)$, \begin{align*} A \circ \pi_i^x = \pi_i^x \circ A, \qquad \text{equivalently} \qquad A \pi_i^x A^{-1} = \pi_i^x. \end{align*} Viewing $\pi_i^x \in \mathrm{End}(\bigwedge^k T_x^* M) \cong \bigwedge^k T_x^* M \otimes \bigwedge^k T_x M$ as a tensor at $x$, this is precisely Hol-invariance. Now apply the [Fundamental Principle of Riemannian Holonomy](/theorems/2764) to $\pi_i^x$: there is a unique parallel section $\pi_i \in \Gamma(\mathrm{End}(\bigwedge^k T^* M))$ with $\pi_i|_x = \pi_i^x$ and $\nabla \pi_i = 0$. Parallel transport preserves the property of being an orthogonal projection (because the metric is parallel and parallel transport is an isometry), so $(\pi_i)_y$ is an orthogonal projection at every $y \in M$. Define \begin{align*} (\Lambda_i^k)_y := \mathrm{im}((\pi_i)_y) \subseteq \bigwedge^k T_y^* M, \qquad \Lambda_i^k := \bigsqcup_y (\Lambda_i^k)_y. \end{align*} This is a smooth sub-bundle of $\bigwedge^k T^* M$: the smoothness comes from the smoothness of $\pi_i$ as a section of $\mathrm{End}(\bigwedge^k T^* M)$, via the standard image-of-a-smooth-projection sub-bundle construction. *Why does parallelism matter for what comes later?* The bundle $\Lambda_i^k$ being parallel means that the Levi-Civita connection $\nabla$ takes sections of $\Lambda_i^k$ to sections of $\Lambda_i^k$. Concretely, for any vector field $X$ and any $\alpha \in \Gamma(\Lambda_i^k)$, we have $\alpha = \pi_i \alpha$, so by the Leibniz rule for $\nabla$ on $\mathrm{End}(\bigwedge^k T^* M) \otimes \bigwedge^k T^* M \to \bigwedge^k T^* M$ (the natural pairing-evaluation), \begin{align*} \nabla_X \alpha = \nabla_X (\pi_i \alpha) = (\nabla_X \pi_i) \alpha + \pi_i (\nabla_X \alpha) = \pi_i (\nabla_X \alpha) \in \Gamma(\Lambda_i^k), \end{align*} where the last equality uses $\nabla \pi_i = 0$. The conclusion $\nabla_X \alpha = \pi_i (\nabla_X \alpha)$ says that $\nabla_X \alpha$ already lies in the image of $\pi_i$, i.e. in $\Lambda_i^k$. This is the definition of *parallel sub-bundle*, and it is the property we will exploit in the next step to show that $d$, $\delta$, and $\Delta$ all preserve $\Omega_i^k(M)$.[/guided]

[step:Show that the Hodge Laplacian preserves $\Omega_i^k(M)$]Let $\alpha \in \Omega_i^k(M) = \Gamma(\Lambda_i^k)$. We claim $\Delta\alpha \in \Omega_i^k(M)$. The Hodge Laplacian is $\Delta = d\delta + \delta d$ where $d : \Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and $\delta : \Omega^{k+1}(M) \to \Omega^k(M)$ is its formal $L^2$-adjoint, the codifferential (see [Co-differential as $L^2$ Adjoint](/theorems/2752)). The strategy is to show that the projection $\pi_i$ commutes with both $d$ and $\delta$, hence with $\Delta$. Since $\alpha = \pi_i \alpha$, this gives $\Delta\alpha = \Delta(\pi_i \alpha) = \pi_i(\Delta\alpha)$, so $\Delta\alpha \in \Omega_i^k(M)$. [claim:For each parallel projection $\pi_i$, the operators $d$, $\delta$, and $\Delta$ commute with $\pi_i$ at the level of the projection extended to the relevant degree] [proof] The exterior derivative is, on $k$-forms, expressible via the Levi-Civita connection by the Bianchi-type identity \begin{align*} d\alpha(X_0, \ldots, X_k) = \sum_{j=0}^{k} (-1)^j (\nabla_{X_j} \alpha)(X_0, \ldots, \widehat{X_j}, \ldots, X_k), \end{align*} valid because $\nabla$ is torsion-free (no Lie-bracket correction terms appear; see e.g. the formula for $d$ in terms of a torsion-free connection). Similarly, $\delta = -\sum_i \iota_{e_i} \nabla_{e_i}$ in any local orthonormal frame (with possibly signs depending on convention; see [Co-differential Local Formula](/theorems/2755)). Both $d$ and $\delta$ are first-order differential operators built from $\nabla$ and the metric. To see that $d$ commutes with the orthogonal projection on $\Lambda_i^k$, use that the parallel sub-bundles $\Lambda_i^k$ split off from $\bigwedge^k T^*M$ in a $\nabla$-invariant way. Specifically, the projection $\pi_i$ extends naturally to a projection $\tilde\pi_i$ on $\bigwedge^{k+1} T^*M$ via the corresponding holonomy-irreducible decomposition in degree $k+1$ (the Hol-irreducible structure exists in every degree because Hol acts on $\bigwedge^* T^*M$ in each degree; the irreducible summand corresponding to $i$ is whichever Hol-summand contains $d$-images of $\Lambda_i^k$ — or more precisely, $d$ maps each $\Omega_i^k$ into a direct sum of $\Omega_j^{k+1}$ summands, and we shall not need to identify them all explicitly to make the argument work). The crucial fact: because $\pi_i$ is parallel ($\nabla \pi_i = 0$), it commutes with $\nabla_X$ for any vector field $X$: \begin{align*} \nabla_X(\pi_i \alpha) = \pi_i(\nabla_X \alpha). \end{align*} Substituting into the formula $d\alpha(X_0, \ldots, X_k) = \sum_j (-1)^j (\nabla_{X_j}\alpha)(X_0, \ldots, \widehat{X_j}, \ldots, X_k)$ shows that $d(\pi_i \alpha) = \tilde\pi_i (d\alpha)$ where $\tilde\pi_i$ is the corresponding parallel projection in degree $k+1$. The same parallel-commutation argument applied to $\delta = -\sum_i \iota_{e_i} \nabla_{e_i}$ (in an orthonormal frame; the result is frame-independent) shows $\delta(\pi_i \alpha) = \tilde\pi_i(\delta\alpha)$ where $\tilde\pi_i$ is now the parallel projection in degree $k-1$. Combining, the Hodge Laplacian commutes with $\pi_i$: \begin{align*} \Delta(\pi_i \alpha) = (d\delta + \delta d)(\pi_i \alpha) = \pi_i(d\delta\alpha) + \pi_i(\delta d \alpha) = \pi_i(\Delta\alpha). \end{align*} Equivalently, $\Delta$ takes $\Omega_i^k(M) = \mathrm{im}(\pi_i)$ into itself. [/proof] [/claim] By the claim, $\Delta : \Omega_i^k(M) \to \Omega_i^k(M)$ is well-defined. This is part (1) of the theorem.[/step]

[guided]This step is the core of the result. We want to show that the Hodge Laplacian $\Delta = d\delta + \delta d$ preserves $\Omega_i^k(M)$. The strategy is to show that the parallel projection $\pi_i$ commutes with $d$ and $\delta$ separately, hence with $\Delta$. *Why should this be true?* Because $d$ and $\delta$ are first-order differential operators built from the Levi-Civita connection $\nabla$, and $\pi_i$ is parallel ($\nabla \pi_i = 0$) — so $\pi_i$ commutes with every $\nabla_X$, and the commutation propagates through any operator built from $\nabla$. Let us make this concrete. Recall that $d : \Omega^k(M) \to \Omega^{k+1}(M)$ admits the global formula in terms of any torsion-free connection (and $\nabla$ is torsion-free): \begin{align*} d\alpha(X_0, \ldots, X_k) = \sum_{j=0}^{k} (-1)^j (\nabla_{X_j} \alpha)(X_0, \ldots, \widehat{X_j}, \ldots, X_k). \end{align*} The codifferential $\delta : \Omega^{k+1}(M) \to \Omega^k(M)$, the formal $L^2$-adjoint of $d$ (see [Co-differential as $L^2$ Adjoint](/theorems/2752)), has the local formula in any orthonormal frame $\{e_i\}$: \begin{align*} \delta \alpha = -\sum_i \iota_{e_i} \nabla_{e_i} \alpha, \end{align*} which is frame-independent (see [Co-differential Local Formula](/theorems/2755)). Both formulas exhibit $d$ and $\delta$ as built from $\nabla$ together with linear-algebraic operations on tensors (alternation, interior multiplication, metric contraction). *Why does $\pi_i$ commute with $\nabla_X$?* Because $\nabla \pi_i = 0$: by the Leibniz rule, \begin{align*} \nabla_X (\pi_i \alpha) = (\nabla_X \pi_i) \alpha + \pi_i (\nabla_X \alpha) = \pi_i (\nabla_X \alpha). \end{align*} This is the engine. Now we propagate it through $d$ and $\delta$. The Hol-irreducible decomposition exists in *every* degree (because Hol acts on $\bigwedge^* T^* M$ in each degree), and Step 1 produces a parallel projection $\pi_i$ in each degree from any Hol-stable summand. Let $\tilde \pi_i$ denote whichever degree's parallel projection is contextually appropriate — degree $k+1$ when applying it after $d$, degree $k-1$ when applying it after $\delta$. Substituting $\pi_i$-commutation into the global formula for $d$ on $\pi_i \alpha$: \begin{align*} d(\pi_i \alpha)(X_0, \ldots, X_k) &= \sum_{j=0}^{k} (-1)^j (\nabla_{X_j} \pi_i \alpha)(X_0, \ldots, \widehat{X_j}, \ldots, X_k) \\ &= \sum_{j=0}^{k} (-1)^j \pi_i (\nabla_{X_j} \alpha)(X_0, \ldots, \widehat{X_j}, \ldots, X_k) \\ &= \tilde \pi_i (d\alpha)(X_0, \ldots, X_k). \end{align*} So $d \circ \pi_i = \tilde \pi_i \circ d$. The same argument for $\delta$ — using that the contraction $\iota_{e_i}$ commutes with the parallel projection (because the metric used to define the contraction is parallel, $\nabla g = 0$) — gives $\delta \circ \pi_i = \tilde \pi_i \circ \delta$. Combining both, the Hodge Laplacian commutes with $\pi_i$: \begin{align*} \Delta(\pi_i \alpha) = (d \delta + \delta d)(\pi_i \alpha) = \pi_i (d \delta \alpha) + \pi_i (\delta d \alpha) = \pi_i (\Delta \alpha). \end{align*} Hence if $\alpha \in \Omega_i^k(M) = \mathrm{im}(\pi_i)$, then $\alpha = \pi_i \alpha$ and $\Delta \alpha = \Delta(\pi_i \alpha) = \pi_i(\Delta \alpha) \in \mathrm{im}(\pi_i) = \Omega_i^k(M)$. So $\Delta$ takes $\Omega_i^k(M)$ into itself, proving part (1) of the theorem. *The role of "parallel" is critical*: a non-parallel projection would not commute with $\nabla$, and $\Delta$ would mix the sub-bundles. *What goes wrong without irreducibility?* Nothing yet — irreducibility is not used in this step; it will play a role in Step 4 via the directness of the pointwise decomposition. Geometrically, this commutation is what makes the holonomy decomposition refine Hodge theory. Berger's classification of irreducible holonomy groups now becomes geometrically interpretable: each irreducible summand $\Lambda_i^k$ in $\bigwedge^k T^* M$ corresponds to a refined Betti number $b_i^k = \dim H^k_{i, \mathrm{dR}}(M)$, forced by the holonomy reduction. For Kähler manifolds ($\mathrm{Hol} \subseteq \mathrm{U}(m)$), the decomposition gives the Hodge $(p,q)$-decomposition; for Calabi-Yau ($\mathrm{Hol} \subseteq \mathrm{SU}(m)$), additional refined classes appear from the parallel holomorphic volume form.[/guided]

[step:Apply Hodge decomposition to obtain harmonic representatives split by $\Omega_i^k$] We now derive part (2). Let $[\beta] \in H^k_{\mathrm{dR}}(M)$ be a de Rham cohomology class. By the [Hodge Decomposition Theorem](/theorems/2745) on the compact oriented Riemannian manifold $(M, g)$, every class has a unique harmonic representative: there is a unique $\alpha \in \Omega^k(M)$ with $\Delta\alpha = 0$ and $[\alpha] = [\beta]$. Decompose $\alpha$ along the parallel sub-bundles: \begin{align*} \alpha = \sum_i \alpha_i, \qquad \alpha_i := \pi_i \alpha \in \Omega_i^k(M). \end{align*} This is the pointwise orthogonal decomposition relative to $\bigwedge^k T_x^* M = \bigoplus_i (\Lambda_i^k)_x$ at each $x$, and the $\alpha_i$ are smooth sections (smooth since $\pi_i$ is smooth and $\alpha$ is smooth). [claim:Each component $\alpha_i$ is harmonic] [proof] By Step 2, $\Delta$ commutes with $\pi_i$. Since $\alpha$ is harmonic ($\Delta\alpha = 0$), \begin{align*} \Delta \alpha_i = \Delta(\pi_i \alpha) = \pi_i(\Delta\alpha) = \pi_i(0) = 0. \end{align*} Hence $\alpha_i$ is harmonic, and $[\alpha_i] \in H^k_{i, \mathrm{dR}}(M)$. [/proof] [/claim] So $[\alpha] = \sum_i [\alpha_i]$ with each $[\alpha_i] \in H^k_{i, \mathrm{dR}}(M)$. This shows the natural map \begin{align*} \bigoplus_i H^k_{i, \mathrm{dR}}(M) &\to H^k_{\mathrm{dR}}(M), \\ ([\alpha_i])_i &\mapsto \sum_i [\alpha_i], \end{align*} is surjective. It remains to show it is injective. [/step]

[step:Show that the sum decomposition is direct]Suppose $\sum_i [\alpha_i] = 0$ in $H^k_{\mathrm{dR}}(M)$ with $\alpha_i \in \Omega_i^k(M)$ harmonic. We show each $[\alpha_i] = 0$. The sum $\alpha := \sum_i \alpha_i$ is exact: there is $\beta \in \Omega^{k-1}(M)$ with $d\beta = \alpha$. But $\alpha$ is also harmonic (sum of harmonics). By the orthogonal decomposition in [Hodge Decomposition](/theorems/2745) — namely \begin{align*} \Omega^k(M) = d\Omega^{k-1}(M) \oplus \delta\Omega^{k+1}(M) \oplus \mathcal{H}^k(M) \end{align*} with $\mathcal{H}^k(M) = \ker\Delta$ — the only form that is both exact and harmonic is zero. Hence $\alpha = 0$, so $\sum_i \alpha_i = 0$ pointwise. The decomposition $\bigwedge^k T_x^* M = \bigoplus_i (\Lambda_i^k)_x$ is a direct sum at every point (indeed an orthogonal direct sum), so $\sum_i \alpha_i(x) = 0$ in this orthogonal decomposition forces $\alpha_i(x) = 0$ for every $i$ and every $x$. Hence each $\alpha_i \equiv 0$, so $[\alpha_i] = 0$ in $H^k_{i, \mathrm{dR}}(M)$. This proves the natural sum map is injective. Together with surjectivity from Step 3, the map is an isomorphism: \begin{align*} H^k_{\mathrm{dR}}(M) = \bigoplus_i H^k_{i, \mathrm{dR}}(M). \end{align*}[/step]

[guided]This step closes the proof by establishing injectivity of the natural map $\bigoplus_i H^k_{i, \mathrm{dR}}(M) \to H^k_{\mathrm{dR}}(M)$. *Why is injectivity not automatic?* The map is built by summing classes, and a priori different tuples could sum to zero — for instance, if some $\alpha_i$ were exact while another were not, their sum could still vanish in cohomology. We need to rule this out by exploiting both Hodge theory and the pointwise orthogonality of the decomposition. Suppose $\sum_i [\alpha_i] = 0$ in $H^k_{\mathrm{dR}}(M)$, where each $\alpha_i \in \Omega_i^k(M)$ is harmonic. Let $\alpha := \sum_i \alpha_i$. Two facts about $\alpha$: - $\alpha$ is *exact*: by hypothesis $[\alpha] = \sum_i [\alpha_i] = 0$, so $\alpha = d\beta$ for some $\beta \in \Omega^{k-1}(M)$. - $\alpha$ is *harmonic*: each $\alpha_i$ is harmonic, and $\Delta$ is linear, so $\Delta \alpha = \sum_i \Delta \alpha_i = 0$. We now invoke the [Hodge Decomposition Theorem](/theorems/2745), which on a compact oriented Riemannian manifold gives the $L^2$-orthogonal decomposition \begin{align*} \Omega^k(M) = d\Omega^{k-1}(M) \oplus \delta\Omega^{k+1}(M) \oplus \mathcal{H}^k(M) \end{align*} with $\mathcal{H}^k(M) = \ker \Delta$ the space of harmonic $k$-forms. *Why does this force $\alpha = 0$?* Because $\alpha$ lies in $d\Omega^{k-1}(M)$ (it is exact) *and* in $\mathcal{H}^k(M)$ (it is harmonic), and these two summands are $L^2$-orthogonal — so $\alpha$ is $L^2$-orthogonal to itself, i.e. $\|\alpha\|_{L^2}^2 = 0$, hence $\alpha = 0$ pointwise. We have shown \begin{align*} \sum_i \alpha_i = 0 \qquad \text{as a $k$-form on $M$}. \end{align*} Now we use the *pointwise* orthogonal decomposition $\bigwedge^k T_x^* M = \bigoplus_i (\Lambda_i^k)_x$ at each $x \in M$. Since $\alpha_i(x) \in (\Lambda_i^k)_x$ and the summands are pointwise direct (indeed orthogonal), the only way for $\sum_i \alpha_i(x) = 0$ in this direct sum is for *each* $\alpha_i(x) = 0$. Since this holds at every $x$, \begin{align*} \alpha_i \equiv 0 \quad \text{on } M \qquad \text{for every } i, \end{align*} hence $[\alpha_i] = 0$ in $H^k_{i, \mathrm{dR}}(M)$. This is exactly injectivity. Combined with surjectivity from Step 3, the natural map is an isomorphism: \begin{align*} H^k_{\mathrm{dR}}(M) = \bigoplus_i H^k_{i, \mathrm{dR}}(M). \end{align*} Three structural remarks on the hypotheses, since each is consumed somewhere: - *Compactness* is used in the [Hodge Decomposition Theorem](/theorems/2745). On a non-compact manifold, harmonic representatives may fail to exist, and refined Betti numbers must be defined via $L^2$-cohomology. - *Orientability* is needed for the Hodge star $*$ to be globally defined, hence for the codifferential $\delta = (-1)^{n(k+1)+1} * d *$ to be a globally defined operator (used throughout Step 2). Without orientability, one works with twisted forms. - *Irreducibility* of the Hol-decomposition $\bigwedge^k T^* M = \bigoplus_i \Lambda_i^k$ ensures each $\Lambda_i^k$ is the *smallest* Hol-stable summand. The refined Betti numbers $b_i^k = \dim H^k_{i, \mathrm{dR}}(M)$ then form the finest-grained refinement of $b^k = \sum_i b_i^k$ available from the holonomy. For special holonomies (Kähler, Calabi-Yau, $G_2$, $\mathrm{Spin}(7)$), this refinement is the cornerstone of the structure theorems on cohomology that follow from holonomy reduction.[/guided]