[proofplan] We work directly from the definitions $\Delta = d\delta + \delta d$ and $\delta = (-1)^{n(k+1) + 1} \star d \star$ on $k$-forms. The proof reduces to two intertwining identities between $\star$ and the operators $d, \delta$: \begin{align*} \star d &= (-1)^{k+1} \delta \star \qquad \text{on } \Omega^k(M), \\ \star \delta &= (-1)^k d \star \qquad \text{on } \Omega^k(M), \end{align*} which both follow from the [Double Hodge Star](/theorems/2740) formula and the definition of $\delta$. Once these two intertwiners are established, $\star \Delta = \star (d\delta + \delta d) = (d\delta + \delta d) \star = \Delta \star$ follows by direct manipulation: each composition of $\star$ with $d$ or $\delta$ flips through both intertwiners, and the four sign factors multiply to $+1$. [/proofplan] [step:Fix conventions and state the two intertwining identities] Let $(M, g)$ be an oriented Riemannian $n$-manifold. On $k$-forms, the codifferential is defined by \begin{align*} \delta : \Omega^k(M) &\to \Omega^{k-1}(M), \\ \beta &\mapsto (-1)^{n(k+1) + 1} \star d \star \beta, \end{align*} and the Hodge Laplacian is \begin{align*} \Delta : \Omega^k(M) &\to \Omega^k(M), \\ \beta &\mapsto d \delta \beta + \delta d \beta. \end{align*} The [Double Hodge Star](/theorems/2740) formula states \begin{align*} \star \star = (-1)^{j(n-j)} \cdot \mathrm{id} \qquad \text{on } \Omega^j(M). \end{align*} Our goal is to prove $\star \Delta = \Delta \star$ as operators $\Omega^k(M) \to \Omega^{n-k}(M)$. The strategy is to first establish two intertwiners, valid on $\Omega^k(M)$: \begin{align*} \star d &= (-1)^{k+1} \delta \star, \tag{I1}\\ \star \delta &= (-1)^k d \star. \tag{I2} \end{align*} These connect $\star$ with the differential and codifferential, and are the algebraic content of the proof. Once available, the commutation $\star \Delta = \Delta \star$ follows mechanically. [/step] [step:Prove the intertwiner $\star d = (-1)^{k+1} \delta \star$ on $\Omega^k(M)$] Let $\alpha \in \Omega^k(M)$. Then $\star \alpha \in \Omega^{n-k}(M)$. Applying $\delta$ at degree $n - k$, with the definition substituted at $k \to n - k$: \begin{align*} \delta (\star \alpha) = (-1)^{n((n-k) + 1) + 1} \star d \star (\star \alpha) = (-1)^{n(n-k+1) + 1} \star d \star \star \alpha. \end{align*} By the [Double Hodge Star](/theorems/2740) formula at degree $k$, \begin{align*} \star \star \alpha = (-1)^{k(n-k)} \alpha. \end{align*} Substituting, \begin{align*} \delta (\star \alpha) = (-1)^{n(n-k+1) + 1} \cdot (-1)^{k(n-k)} \cdot \star d \alpha. \end{align*} Compute the combined sign exponent modulo $2$: \begin{align*} n(n-k+1) + 1 + k(n-k) &= n^2 - nk + n + 1 + kn - k^2 \\ &= n^2 + n + 1 - k^2 \\ &\equiv (n^2 + n) + 1 + k^2 \pmod 2 \\ &\equiv 0 + 1 + k \pmod 2 \\ &\equiv k + 1 \pmod 2, \end{align*} where we used $n^2 + n = n(n+1) \equiv 0 \pmod 2$ (consecutive integers) and $k^2 \equiv k \pmod 2$. Therefore \begin{align*} \delta(\star \alpha) = (-1)^{k+1} \star d \alpha, \end{align*} which rearranges to \begin{align*} \star d \alpha = (-1)^{k+1} \delta (\star \alpha) \qquad \text{for all } \alpha \in \Omega^k(M). \end{align*} This is the intertwiner (I1). [guided] We want to relate $\star d$ to $\delta \star$ on $\Omega^k(M)$. The strategy is to work backwards: rather than starting from $\star d \alpha$ and trying to manufacture a $\delta \star$ on the right, we start from $\delta(\star \alpha)$ — which is just the codifferential applied to a form whose degree we know — and unfold the definition of $\delta$ until $\star d \alpha$ appears on the right. Let $\alpha \in \Omega^k(M)$, so $\star \alpha \in \Omega^{n-k}(M)$. The codifferential $\delta$ on a $j$-form is $(-1)^{n(j+1) + 1} \star d \star$, so substituting $j = n - k$ gives \begin{align*} \delta (\star \alpha) = (-1)^{n((n-k) + 1) + 1} \star d \star (\star \alpha) = (-1)^{n(n-k+1) + 1} \star d \star \star \alpha. \end{align*} The double star $\star \star$ acts on $\alpha \in \Omega^k(M)$, so we apply the [Double Hodge Star](/theorems/2740) formula at degree $j = k$: \begin{align*} \star \star \alpha = (-1)^{k(n-k)} \alpha. \end{align*} This collapses the chain of stars and replaces them with a sign: \begin{align*} \delta (\star \alpha) = (-1)^{n(n-k+1) + 1} \cdot (-1)^{k(n-k)} \cdot \star d \alpha. \end{align*} At this point $\star d \alpha$ has appeared, and we just need to simplify the exponent. The combined sign exponent expands as \begin{align*} n(n-k+1) + 1 + k(n-k) &= n^2 - nk + n + 1 + kn - k^2 \\ &= n^2 + n + 1 - k^2. \end{align*} We now reduce this modulo $2$. Two parity facts handle every term: 1. $n^2 + n = n(n+1)$ is the product of two consecutive integers, so one of them is even and the product is even: $n^2 + n \equiv 0 \pmod 2$. 2. $k^2 \equiv k \pmod 2$ (squaring preserves parity: $0^2 = 0$, $1^2 = 1$). These same mod-$2$ identities pervade the sign bookkeeping in Hodge theory; the same trick appears in the proof of the [Co-differential is Formal Adjoint of $d$](/theorems/2742). Applying both: \begin{align*} n^2 + n + 1 - k^2 \equiv 0 + 1 + k \equiv k + 1 \pmod 2, \end{align*} where the minus sign is irrelevant since $-k^2 \equiv k^2 \pmod 2$. Therefore \begin{align*} \delta(\star \alpha) = (-1)^{k+1} \star d \alpha. \end{align*} Multiplying both sides by $(-1)^{k+1}$ (an involution) rearranges this into the desired form: \begin{align*} \star d \alpha = (-1)^{k+1} \delta (\star \alpha) \qquad \text{for all } \alpha \in \Omega^k(M). \end{align*} This is the intertwiner (I1). The geometric content of (I1) is that $\star$ "intertwines" $d$ on $\Omega^k$ with $\delta$ on $\Omega^{n-k}$: applying $\star$ to a $k$-form and then taking the codifferential is the same (up to a sign) as taking the differential and then applying $\star$. This is the source of the symmetry between de Rham cohomology in degrees $k$ and $n - k$ on a closed oriented manifold (Poincaré duality, in its differential-form realisation). [/guided] [/step] [step:Prove the intertwiner $\star \delta = (-1)^k d \star$ on $\Omega^k(M)$] Let $\beta \in \Omega^k(M)$. By the definition of $\delta$, \begin{align*} \star \delta \beta = \star \big[ (-1)^{n(k+1) + 1} \star d \star \beta \big] = (-1)^{n(k+1) + 1} \cdot \star \star (d \star \beta). \end{align*} The form $d \star \beta$ has degree $(n - k) + 1 = n - k + 1$. Applying the [Double Hodge Star](/theorems/2740) formula at degree $j = n - k + 1$: \begin{align*} \star \star (d \star \beta) = (-1)^{(n-k+1)(n - (n-k+1))} \cdot d \star \beta = (-1)^{(n-k+1)(k-1)} \cdot d \star \beta. \end{align*} Substituting, \begin{align*} \star \delta \beta = (-1)^{n(k+1) + 1} \cdot (-1)^{(n-k+1)(k-1)} \cdot d \star \beta. \end{align*} Compute the combined sign exponent modulo $2$: \begin{align*} n(k+1) + 1 + (n-k+1)(k-1) &= nk + n + 1 + nk - n - k^2 + k + k - 1 \\ &= 2nk - k^2 + 2k \\ &\equiv -k^2 \pmod 2 \\ &\equiv k^2 \pmod 2 \\ &\equiv k \pmod 2, \end{align*} using $2nk \equiv 0$, $2k \equiv 0$, and $k^2 \equiv k \pmod 2$. Therefore \begin{align*} \star \delta \beta = (-1)^k \cdot d \star \beta \qquad \text{for all } \beta \in \Omega^k(M). \end{align*} This is the intertwiner (I2). [/step] [step:Apply (I1) and (I2) to compute $\star \Delta$ in terms of $\Delta \star$] Let $\alpha \in \Omega^k(M)$. By the definition of the Laplacian and linearity of $\star$, \begin{align*} \star \Delta \alpha = \star d \delta \alpha + \star \delta d \alpha. \end{align*} We treat each of the two summands. **First summand $\star d \delta \alpha$.** The form $\delta \alpha$ has degree $k - 1$. Apply (I1) at degree $k - 1$ to the inner $\star d$: \begin{align*} \star d (\delta \alpha) = (-1)^{(k-1)+1} \delta \star (\delta \alpha) = (-1)^k \delta (\star \delta \alpha). \end{align*} Now apply (I2) at degree $k$ to the inner $\star \delta$: \begin{align*} \star \delta \alpha = (-1)^k d \star \alpha. \end{align*} Substituting, \begin{align*} \star d \delta \alpha = (-1)^k \delta \big[(-1)^k d \star \alpha\big] = (-1)^k \cdot (-1)^k \cdot \delta d (\star \alpha) = \delta d (\star \alpha). \end{align*} **Second summand $\star \delta d \alpha$.** The form $d \alpha$ has degree $k + 1$. Apply (I2) at degree $k + 1$ to the inner $\star \delta$: \begin{align*} \star \delta (d \alpha) = (-1)^{k+1} d \star (d \alpha) = (-1)^{k+1} d (\star d \alpha). \end{align*} Now apply (I1) at degree $k$ to the inner $\star d$: \begin{align*} \star d \alpha = (-1)^{k+1} \delta \star \alpha. \end{align*} Substituting, \begin{align*} \star \delta d \alpha = (-1)^{k+1} d \big[(-1)^{k+1} \delta \star \alpha\big] = (-1)^{k+1} \cdot (-1)^{k+1} \cdot d \delta (\star \alpha) = d \delta (\star \alpha). \end{align*} [guided] The strategy is to show that each of the two summands of $\star \Delta \alpha$ converts into its structural mirror: $\star d \delta$ should become $\delta d \star$, and $\star \delta d$ should become $d \delta \star$. Each conversion uses the two intertwiners (I1) and (I2) in succession — once each — and the question is whether the resulting sign factors cancel. We will see that they do, but only because $\Delta = d\delta + \delta d$ is *symmetric* in $d$ and $\delta$. Let $\alpha \in \Omega^k(M)$. By the definition $\Delta = d\delta + \delta d$ and linearity of $\star$: \begin{align*} \star \Delta \alpha = \star d \delta \alpha + \star \delta d \alpha. \end{align*} We treat each summand in turn. **First summand $\star d \delta \alpha$.** The form $\delta \alpha$ has degree $k - 1$, so to apply (I1) — which acts on a $k$-form — we substitute $k \to k - 1$: \begin{align*} \star d (\delta \alpha) = (-1)^{(k-1)+1} \delta \star (\delta \alpha) = (-1)^k \delta (\star \delta \alpha). \end{align*} Now $\star \delta \alpha$ appears, with $\alpha$ at its original degree $k$. Apply (I2) at degree $k$: \begin{align*} \star \delta \alpha = (-1)^k d \star \alpha. \end{align*} Substituting this into the previous line: \begin{align*} \star d \delta \alpha = (-1)^k \delta \big[(-1)^k d \star \alpha\big] = (-1)^k \cdot (-1)^k \cdot \delta d (\star \alpha) = \delta d (\star \alpha). \end{align*} The two sign factors $(-1)^k$ multiply to $+1$. This is the first miracle of the proof: the inner application of (I2) at degree $k$ produces sign $(-1)^k$, and the outer application of (I1) at degree $k - 1$ also produces sign $(-1)^k$ (because (I1) at degree $j$ has sign $(-1)^{j+1}$, and $j = k - 1$ gives $(-1)^k$). The signs are equal, so they square to $+1$. **Second summand $\star \delta d \alpha$.** The form $d \alpha$ has degree $k + 1$, so to apply (I2) — which acts on a $k$-form — we substitute $k \to k + 1$: \begin{align*} \star \delta (d \alpha) = (-1)^{k+1} d \star (d \alpha) = (-1)^{k+1} d (\star d \alpha). \end{align*} Now $\star d \alpha$ appears, with $\alpha$ at its original degree $k$. Apply (I1) at degree $k$: \begin{align*} \star d \alpha = (-1)^{k+1} \delta \star \alpha. \end{align*} Substituting this into the previous line: \begin{align*} \star \delta d \alpha = (-1)^{k+1} d \big[(-1)^{k+1} \delta \star \alpha\big] = (-1)^{k+1} \cdot (-1)^{k+1} \cdot d \delta (\star \alpha) = d \delta (\star \alpha). \end{align*} Again the two sign factors are equal — both $(-1)^{k+1}$ — and they square to $+1$. Why do the signs cancel pairwise in both summands? In the first summand, both intertwiners act at "shifted" degree $k - 1$ on the inner step and "unshifted" degree $k$ on the outer, but the sign formulas $(-1)^{j+1}$ for (I1) and $(-1)^j$ for (I2) shift in opposite ways under $j \mapsto j - 1$, so the two signs end up agreeing. In the second summand, the same arithmetic plays out shifted by one, again producing a matching pair. This is the algebraic reason that $\star$ commutes with the *symmetric* combination $d\delta + \delta d$ but not, for example, with $d\delta - \delta d$ — the asymmetric combination would acquire a leftover sign. This is the algebraic shadow of a deeper fact: $\Delta$ is the unique second-order differential operator on $\Omega^*(M)$ that commutes with $\star$ and is a "Laplacian" in the sense of arising from $d$ and its adjoint. The Bochner–Weitzenböck formula provides another expression for $\Delta$ in terms of the connection Laplacian and the curvature, and that formula too commutes with $\star$ — for the same algebraic reasons. [/guided] [/step] [step:Combine the two summands to conclude $\star \Delta = \Delta \star$] From the previous step, \begin{align*} \star \Delta \alpha = \star d \delta \alpha + \star \delta d \alpha = \delta d (\star \alpha) + d \delta (\star \alpha) = (d \delta + \delta d)(\star \alpha) = \Delta (\star \alpha). \end{align*} Since this holds for every $\alpha \in \Omega^k(M)$ and every $k \in \{0, 1, \ldots, n\}$, \begin{align*} \star \Delta = \Delta \star \qquad \text{as operators on } \Omega^*(M). \end{align*} This completes the proof. [/step]