= Solution
Use the given orientation throughout, and the usual convention that the manifold has no boundary. In an oriented <manifold chart> the <metric volume form> is
$$
\omega_g=\sqrt{\det(g_{ij})}\,dx^1\wedge\cdots\wedge dx^n.
$$
If $y$ is another oriented chart and $A=\partial x/\partial y$, then $g_y=A^tg_xA$, so $\sqrt{\det g_y}=(\det A)\sqrt{\det g_x}$ because $\det A>0$. This is exactly the transformation of the top <exterior product>. The formulas therefore agree on overlaps and define a smooth positive <volume form>. Equivalently it takes value one on any positively oriented orthonormal tangent frame.
The metric induces an <inner product> on $p$-forms by making the increasing exterior products of an orthonormal coframe orthonormal. The <Hodge star operator> is the unique pointwise linear map $*:\Lambda^pT^*M\to\Lambda^{n-p}T^*M$ satisfying
$$
\eta\wedge *\theta=\langle\eta,\theta\rangle_g\omega_g.
$$
For an increasing multi-index $I$, $*e^I$ is the complementary wedge with the sign making $e^I\wedge *e^I=\omega_g$. Swapping the blocks of $p$ and $n-p$ factors introduces $(-1)^{p(n-p)}$. Therefore
$$
\boxed{*^2=(-1)^{p(n-p)}\operatorname{id}\quad\text{on }\Omega^p(M).}
$$
In particular $*1=\omega_g$ and $*\omega_g=1$.
For compactly supported smooth $f$, the $(n-1)$-form $f*\alpha$ has compact support. <Stokes theorem> and the <graded Leibniz rule> give
$$
0=\int_Md(f*\alpha)=\int_Mdf\wedge *\alpha+\int_Mf\,d*\alpha.
$$
A top form $\tau$ equals $(*\tau)\omega_g$. Consequently this <Hodge integration by parts for one-forms> becomes exactly
$$
\int_M(-f*d*\alpha)\omega_g=\int_M\langle df,\alpha\rangle_g\omega_g.
$$
Compact support of $f$ is enough; $\alpha$ need not itself have compact support.
Define the <codifferential> on $p$-forms by $\delta=(-1)^{n(p+1)+1}*d*$ and the <Hodge Laplacian> by $\Delta=d\delta+\delta d$. A <harmonic differential form> is a smooth form in $\ker\Delta$. On functions this is the <positive Laplace-Beltrami operator>, $\Delta f=\delta df=-\operatorname{div}_g\operatorname{grad}_gf$. This sign convention is required by the product identity; it is the negative of the $\operatorname{div}\operatorname{grad}$ convention also commonly used for the <Laplace-Beltrami operator>.
For a $p$-form $\beta$, the square of the <Hodge star operator> and the <codifferential> formula give
$$
\delta(*\beta)=(-1)^{p+1}*d\beta,\qquad d(*\beta)=(-1)^p*\delta\beta.
$$
Applying the second formula to $d\beta$ and the first to $\delta\beta$ gives
$$
\Delta(*\beta)=(-1)^{p+1}d(*d\beta)+(-1)^p\delta(*\delta\beta)=*\delta d\beta+*d\delta\beta=*\Delta\beta.
$$
Thus <Hodge star commutes with the Hodge Laplacian>. Since $*$ is invertible, \b[$\beta$ is harmonic if and only if $*\beta$ is harmonic.] This holds without compactness; we have not used the generally false noncompact implication that a harmonic form must be closed and coclosed.
For a smooth function $h$ and one-form $\alpha$, the same <graded Leibniz rule> gives $\delta(h\alpha)=h\delta\alpha-*\,(dh\wedge *\alpha)=h\delta\alpha-\langle dh,\alpha\rangle_g$. Apply this to $d(f_1f_2)=f_2df_1+f_1df_2$ to obtain the <product rule for the positive Laplace-Beltrami operator>
$$
\boxed{\Delta(f_1f_2)=f_2\Delta f_1+f_1\Delta f_2-2\langle df_1,df_2\rangle_g.}
$$
The <Hodge decomposition theorem> states that on a compact oriented boundaryless <Riemannian manifold>, smooth forms have the $L^2$-orthogonal decomposition
$$
\Omega^p(M)=\mathcal H^p(M)\oplus d\Omega^{p-1}(M)\oplus\delta\Omega^{p+1}(M),\qquad\mathcal H^p(M)=\ker\Delta,
$$
and each <de Rham cohomology> class has a unique <harmonic differential form> representative. To spell out the last conclusion, a harmonic form is closed and coclosed because $\langle\Delta\eta,\eta\rangle=\|d\eta\|^2+\|\delta\eta\|^2$. If a closed form decomposes as $h+du+\delta v$, then $d\delta v=0$; integration by parts gives $\|\delta v\|^2=\langle v,d\delta v\rangle=0$. Thus it represents $h$. A harmonic exact form has zero norm by adjointness, proving uniqueness. The analytic existence of the decomposition is the stated <Hodge decomposition theorem>.
Now let $M$ be compact, connected and oriented. A harmonic function satisfies $0=\langle f,\Delta f\rangle=\|df\|^2$, so it is constant. The <Hodge star operator> identifies $\mathcal H^0(M)$ with $\mathcal H^n(M)$, hence $\mathcal H^n(M)=\mathbb R\omega_g$. A <Riemannian metric> exists by Question 3 even if none was initially chosen. Using the harmonic representative of each class, we obtain the <top de Rham cohomology of a compact connected oriented manifold>
$$
\boxed{H^n_{\mathrm{dR}}(M)\cong\mathbb R,\qquad[\omega_g]\text{ spans it}.}
$$
The class is nonzero also directly from <Stokes theorem>, since $\int_M\omega_g>0$ while every exact top form has zero integral. Boundarylessness matters: a compact interval has zero first <de Rham cohomology>.
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