= Solution
The <exterior derivative> is the real-linear degree-one map $d:\Omega^r(M)\to\Omega^{r+1}(M)$ characterized by $df(X)=Xf$ on <smooth functions>, the graded <Leibniz rule>
$$
d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^r\alpha\wedge d\beta
\qquad(\alpha\in\Omega^r(M)),
$$
and $d^2=0$. In <local coordinates> $(x^1,\ldots,x^m)$, write $\alpha=\sum_I\alpha_I dx^{i_1}\wedge\cdots\wedge dx^{i_r}$ using <multi-index notation>. Since $dx^i=d(x^i)$ and hence $d(dx^i)=0$, the defining rules force
$$
\boxed{d\alpha=\sum_{I,j}\frac{\partial\alpha_I}{\partial x^j}dx^j\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r}.}
$$
This proves local uniqueness, and the coordinate formulas agree on overlaps because the same rules are preserved by the <pullback of a differential form>. They also directly define an operator satisfying all three rules, proving existence.
An <exact differential form> is a form $\omega=d\eta$. Let $\pi:S^{2n}\to\mathbb{RP}^{2n}$ be the antipodal double <covering map> and let $a(x)=-x$. For a $2n$-form $\omega$ on real projective space, $a^*\pi^*\omega=\pi^*\omega$, while the <mapping degree> of the <antipodal map> is $-1$. Therefore
$$
\int_{S^{2n}}\pi^*\omega
=\int_{S^{2n}}a^*\pi^*\omega
=-\int_{S^{2n}}\pi^*\omega=0.
$$
By the stated criterion, $\pi^*\omega=d\eta$. The <invariant primitive under a finite group action>
$$
\bar\eta=\frac12(\eta+a^*\eta)
$$
still satisfies $d\bar\eta=\pi^*\omega$ and descends to a form $\beta$ on $\mathbb{RP}^{2n}$. Since pullback through a covering is injective on differential forms, $\pi^*(d\beta-\omega)=0$ implies $d\beta=\omega$. Thus \b[every $2n$-form on $\mathbb{RP}^{2n}$ is exact]; this is the <top-degree differential forms on even-dimensional real projective space are exact> result.
The $k$th <de Rham cohomology> is
$$
H^k_{\mathrm{dR}}(M)=\frac{\ker(d:\Omega^k(M)\to\Omega^{k+1}(M))}{\operatorname{im}(d:\Omega^{k-1}(M)\to\Omega^k(M))},
$$
the <closed differential forms> modulo the exact ones. For the product, let $p:M\times S^1\to M$ be projection and choose a closed one-form $\nu$ on the circle with $\int_{S^1}\nu=1$. <Averaging differential forms over the circle> is cochain-homotopic to the identity, so every class has a rotation-invariant representative; if such a representative is exact, averaging a primitive gives an invariant primitive. Every invariant $k$-form has a unique decomposition $p^*\alpha+p^*\beta\wedge\nu$, and
$$
d(p^*\alpha+p^*\beta\wedge\nu)=p^*(d\alpha)+p^*(d\beta)\wedge\nu.
$$
Closedness and exactness are therefore componentwise. Hence the map
$$
\boxed{([\alpha],[\beta])\longmapsto[p^*\alpha+p^*\beta\wedge\nu]}
$$
is a well-defined bijection $H^k_{\mathrm{dR}}(M)\oplus H^{k-1}_{\mathrm{dR}}(M)\to H^k_{\mathrm{dR}}(M\times S^1)$, proving the <de Rham cohomology of a product with a circle> formula.
Back to article page
