[proofplan] The proof is the standard energy identity for the Hodge Laplacian. We pair $\Delta_d\alpha$ with $\alpha$ in the Hermitian $L^2$ [inner product](/page/Inner%20Product) and use the defining formal-adjoint relation between $d$ and $d^*$ to rewrite the result as the sum of two squared norms. If $\Delta_d\alpha=0$, this sum is zero, so both terms vanish. Conversely, if $d\alpha$ and $d^*\alpha$ vanish, then the defining formula $\Delta_d=d^*d+dd^*$ gives $\Delta_d\alpha=0$. [/proofplan]

[step:Compute the Hodge Laplacian energy identity]Let $(\cdot,\cdot)_{L^2}$ denote the Hermitian $L^2$ inner product on complex-valued forms induced by $g$, linear in the first argument and conjugate-linear in the second. Thus, for smooth complex-valued $r$-forms $\beta,\gamma\in A^r(M;\mathbb C)$, \begin{align*} (\beta,\gamma)_{L^2}:=\int_M h_g(\beta,\gamma)(x)\,dV_g(x), \end{align*} where $h_g$ is the pointwise Hermitian inner product on $\Lambda^rT^*M\otimes_{\mathbb R}\mathbb C$ and $dV_g$ is the Riemannian volume measure. Since $M$ is compact and has no boundary, the formal adjoint $d^*$ is characterized by \begin{align*} (d\eta,\theta)_{L^2}=(\eta,d^*\theta)_{L^2} \end{align*} for every pair of smooth forms $\eta$ and $\theta$ of compatible degrees. Applying this identity first with $\eta=\alpha$ and $\theta=d\alpha$, and then with $\eta=d^*\alpha$ and $\theta=\alpha$, gives \begin{align*} (\alpha,d^*d\alpha)_{L^2}=(d\alpha,d\alpha)_{L^2}. \end{align*} Also, \begin{align*} (d^*\alpha,d^*\alpha)_{L^2}=(dd^*\alpha,\alpha)_{L^2}. \end{align*} Because $(d^*\alpha,d^*\alpha)_{L^2}$ is real, this last equality implies \begin{align*} (\alpha,dd^*\alpha)_{L^2}=(d^*\alpha,d^*\alpha)_{L^2}. \end{align*} Therefore, using $\Delta_d=d^*d+dd^*$, \begin{align*} (\alpha,\Delta_d\alpha)_{L^2}=\|d\alpha\|_{L^2}^2+\|d^*\alpha\|_{L^2}^2. \end{align*}[/step]

[guided]We want to turn the equation $\Delta_d\alpha=0$ into pointwise information about the two first-order quantities $d\alpha$ and $d^*\alpha$. The mechanism is the $L^2$ energy identity, which says that the Laplacian pairing is exactly a sum of squared norms. Let $(\cdot,\cdot)_{L^2}$ be the Hermitian $L^2$ inner product on complex-valued forms, linear in the first argument and conjugate-linear in the second. For smooth complex-valued $r$-forms $\beta,\gamma\in A^r(M;\mathbb C)$, it is \begin{align*} (\beta,\gamma)_{L^2}:=\int_M h_g(\beta,\gamma)(x)\,dV_g(x), \end{align*} where $h_g$ is the pointwise Hermitian inner product induced by the Riemannian metric and $dV_g$ is the Riemannian volume measure. The definition of $d^*$ as the formal adjoint of $d$ gives \begin{align*} (d\eta,\theta)_{L^2}=(\eta,d^*\theta)_{L^2} \end{align*} whenever the degrees match. The hypotheses that $M$ is compact and has no boundary ensure that this adjoint identity has no boundary term. Apply the adjoint identity with $\eta=\alpha$ and $\theta=d\alpha$. Then \begin{align*} (d\alpha,d\alpha)_{L^2}=(\alpha,d^*d\alpha)_{L^2}. \end{align*} This accounts for the $d^*d$ part of $\Delta_d$. Next apply the same adjoint identity with $\eta=d^*\alpha$ and $\theta=\alpha$. This gives \begin{align*} (dd^*\alpha,\alpha)_{L^2}=(d^*\alpha,d^*\alpha)_{L^2}. \end{align*} Because $(d^*\alpha,d^*\alpha)_{L^2}$ is a squared norm, it is a real number. Taking Hermitian symmetry into account, this is equivalent to \begin{align*} (\alpha,dd^*\alpha)_{L^2}=(d^*\alpha,d^*\alpha)_{L^2}. \end{align*} Adding the two identities and using the definition \begin{align*} \Delta_d=d^*d+dd^* \end{align*} yields \begin{align*} (\alpha,\Delta_d\alpha)_{L^2}=\|d\alpha\|_{L^2}^2+\|d^*\alpha\|_{L^2}^2. \end{align*} This is the desired energy identity.[/guided]

[step:Deduce closedness and coclosedness from harmonicity] Assume that $\alpha$ is $\Delta_d$-harmonic, so $\Delta_d\alpha=0$. Substituting this into the energy identity gives \begin{align*} 0=(\alpha,\Delta_d\alpha)_{L^2}=\|d\alpha\|_{L^2}^2+\|d^*\alpha\|_{L^2}^2. \end{align*} Both summands are nonnegative [real numbers](/page/Real%20Numbers). Hence \begin{align*} \|d\alpha\|_{L^2}^2=0, \qquad \|d^*\alpha\|_{L^2}^2=0. \end{align*} Since $d\alpha$ and $d^*\alpha$ are smooth forms, zero $L^2$ norm implies that they vanish identically. Therefore \begin{align*} d\alpha=0, \qquad d^*\alpha=0. \end{align*} [/step]

[step:Use closedness and coclosedness to recover harmonicity] Conversely, assume that \begin{align*} d\alpha=0, \qquad d^*\alpha=0. \end{align*} Using the defining formula for the Hodge Laplacian, \begin{align*} \Delta_d\alpha=d^*d\alpha+dd^*\alpha. \end{align*} The first term is $d^*(0)=0$, and the second term is $d(0)=0$. Hence \begin{align*} \Delta_d\alpha=0. \end{align*} Thus $\alpha$ is $\Delta_d$-harmonic. Combining the two implications proves the equivalence. [/step]