= Solution
The <exterior derivative> is the real-linear degree-one operator $d:\Omega^k\to\Omega^{k+1}$ characterized by its usual derivative on functions, the <graded Leibniz rule>
$$
d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta\quad(\alpha\in\Omega^k),
$$
and $d^2=0$. In local coordinates,
$$
d\left(\sum_I f_I\,dx^{i_1}\wedge\cdots\wedge dx^{i_k}\right)
=\sum_I df_I\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}.
$$
These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with <pullback of a differential form>.
On the unit ball let $V=x^i\partial_i$ be the radial <vector field>, and let $F_t(x)=tx$. For $k\ge1$ define the <radial homotopy operator> by
$$
(h_k\omega)_x(v_1,\ldots,v_{k-1})
=\int_0^1 t^{k-1}\omega_{tx}(x,v_1,\ldots,v_{k-1})\,dt.
$$
Here the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega=\int_0^1t^{-1}F_t^*(\iota_V\omega)\,dt$, where $\iota_V$ is the <interior product>. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to $\Omega^{k-1}$; set $h_0=0$.
For $t>0$, differentiation of the coordinate pullback and <Cartan's magic formula> give
$$
\frac{d}{dt}F_t^*\omega
=t^{-1}F_t^*\mathcal L_V\omega
=t^{-1}F_t^*(d\iota_V\omega+\iota_Vd\omega).
$$
For completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $dx^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one proves
$$
\boxed{dh_k\omega+h_{k+1}d\omega=\omega-F_0^*\omega.}
$$
For positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega=0$ and $k\ge1$, it follows that $\omega=d(h_k\omega)$. A chart about any point contains such a ball, proving the <Poincaré lemma>. In degree zero the formula is instead $h_1df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This <degree-zero correction to radial homotopy> specifies the necessary positive-degree convention in the printed request.
To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to $\mathbb R^2$, where the same star-shaped homotopy applies. A <closed differential one-form> $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. Thus
$$
\boxed{\alpha=df\text{ on }S^2,\qquad H^1_{\mathrm{dR}}(S^2)=0.}
$$
This proves directly that <closed one-forms on the two-sphere are exact>.
For the <Hodge decomposition theorem>, assume a compact oriented <Riemannian manifold> without boundary. The <Hodge star operator> is characterized by $\alpha\wedge*\beta=\langle\alpha,\beta\rangle\operatorname{vol}_g$ and obeys $**\alpha=(-1)^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta=(-1)^{n(k+1)+1}*d*$. Define the <Hodge Laplacian> $\Delta=d\delta+\delta d$ and the space $\mathcal H^k=\ker\Delta$ of <harmonic differential forms>. Integration by parts, with no boundary term, gives
$$
\langle\Delta\alpha,\alpha\rangle_{L^2}
=\|d\alpha\|_{L^2}^2+\|\delta\alpha\|_{L^2}^2.
$$
Hence a form is harmonic exactly when it is both closed and coclosed. The <Hodge decomposition theorem> states that $\mathcal H^k$ is finite-dimensional and that there is the orthogonal decomposition
$$
\Omega^k(M)=\mathcal H^k\oplus d\Omega^{k-1}(M)\oplus\delta\Omega^{k+1}(M).
$$
If a closed form has decomposition $\alpha=h+d\beta+\delta\gamma$, then $\langle\alpha,\delta\gamma\rangle=\langle d\alpha,\gamma\rangle=0$. Orthogonality consequently forces $\delta\gamma=0$, so its <de Rham cohomology> class is represented by $h$. An exact harmonic form $h=d\beta$ has $\|h\|^2=\langle\delta h,\beta\rangle=0$, proving uniqueness. Thus
$$
\boxed{\mathcal H^k(M)\cong H^k_{\mathrm{dR}}(M).}
$$
Every top-degree form is $\omega=f\operatorname{vol}_g$, and $d\omega=0$ for dimensional reasons. Its <Hodge star> is $f$, so the formula for $\delta$ shows that $\delta\omega=0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the <harmonic top-degree forms on a closed oriented manifold> form the space
$$
\boxed{\mathcal H^n(M)=\mathbb R\,\operatorname{vol}_g,\qquad\dim\mathcal H^n(M)=1.}
$$
Any such smooth manifold admits a <Riemannian metric>; the assertion is valid for every choice of that metric.
Back to article page
