= Solution
On an oriented Riemannian $n$-manifold, the <metric volume form> is the unique top form taking value one on every positively oriented orthonormal tangent basis. In a positive coordinate chart it is
$$
\boxed{\operatorname{vol}_g=\sqrt{\det(g_{ij})}\,dx^1\wedge\cdots\wedge dx^n}.
$$
For another positive coordinate chart $y$, write $A=\partial x/\partial y$. Then $g_y=A^{\mathsf T}g_xA$, so $\sqrt{\det g_y}=\det A\sqrt{\det g_x}$, since $\det A>0$. The wedge of coordinate differentials acquires the same <determinant> factor. The two local expressions agree, proving coordinate independence and smooth global existence. Positivity of the chosen charts is what fixes the sign of the <volume form>.
The metric induces the <inner product on exterior powers of a cotangent space>. Define the <Hodge star operator> $*:\Lambda^pT^*M\to\Lambda^{n-p}T^*M$ by
$$
\boxed{\alpha\wedge*\beta=\langle\alpha,\beta\rangle_g\operatorname{vol}_g}
$$
for forms of the same degree, and apply it fiberwise to smooth forms. In an oriented orthonormal coframe, if $I$ is an increasing $p$-index and $I^c$ its ordered complement, $*e^I=\epsilon(I,I^c)e^{I^c}$. Exchanging the two blocks has sign $(-1)^{p(n-p)}$, so
$$
\boxed{*^2|_{\Omega^p}=(-1)^{p(n-p)}I}.
$$
Use the real $L^2$ <inner product> $(\alpha,\beta)=\int_M\alpha\wedge*\beta$. For a $p$-form $\alpha$ and a $(p+1)$-form $\beta$, the <graded Leibniz rule> and the given boundaryless Stokes identity give
$$
0=\int_M d(\alpha\wedge*\beta)
=\int_M d\alpha\wedge*\beta+(-1)^p\int_M\alpha\wedge d*\beta.
$$
The <codifferential> acting on $\beta$ has sign $(-1)^{n(p+2)+1}$. Applying the star once more and using its square on the $(n-p)$-form $d*\beta$ gives
$$
*\delta\beta=(-1)^{n(p+2)+1+p(n-p)}d*\beta=(-1)^{p+1}d*\beta.
$$
The exponent reduction uses parity, since $p^2\equiv p\pmod2$. Combining the two displayed identities proves $(d\alpha,\beta)=(\alpha,\delta\beta)$. Symmetry of the real <inner product> supplies precisely
$$
\boxed{\int_M\beta\wedge*d\alpha=\int_M\delta\beta\wedge*\alpha}.
$$
Thus <Hodge integration by parts in arbitrary degree> identifies $\delta$ as the formal adjoint of $d$. The displayed bilinear identity extends to complex <differential forms> by complex linearity. For positive norm identities on complex forms, use the Hermitian <inner product> with conjugation instead.
With the nonnegative convention, the <Hodge Laplacian> is
$$
\boxed{\Delta=d\delta+\delta d}.
$$
It is the form-valued extension of the <Laplace-Beltrami operator>, whose action on functions is $-\operatorname{div}\operatorname{grad}$. A <harmonic differential form> is a smooth form in $\ker\Delta$. The adjoint identity gives
$$
(\Delta\alpha,\alpha)=\|d\alpha\|_{L^2}^2+\|\delta\alpha\|_{L^2}^2.
$$
If $\Delta\alpha=0$, both norms vanish, and smoothness implies
$$
\boxed{d\alpha=0,\qquad\delta\alpha=0}.
$$
The closed compact hypotheses are essential for the absence of boundary terms and this global energy conclusion.
The <Hodge decomposition theorem> states that on a closed oriented <Riemannian manifold>
$$
\boxed{\Omega^p(M)=\mathcal H^p(M)\oplus d\Omega^{p-1}(M)\oplus\delta\Omega^{p+1}(M)}
$$
is an $L^2$-orthogonal direct sum, with finite-dimensional harmonic space $\mathcal H^p$. Every smooth form has unique harmonic, exact and coexact components, although primitives of the latter two need not be unique. Equivalently every <de Rham cohomology> class has a unique harmonic representative. An exact harmonic form belongs to both the harmonic and exact summands; their orthogonality makes its squared norm zero. Therefore $\boxed{\alpha\text{ exact and harmonic}\Longrightarrow\alpha=0}$. The same conclusion follows directly from $\alpha=d\eta$ and $\|\alpha\|^2=(\delta\alpha,\eta)=0$.
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