[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*}
[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]
[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]
