Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 115 / 1

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 ruleand $d^2=0$. In local coordinates $(x^1,\dots ,x^m)$, write $\alpha ={\sum }_I{\alpha }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$ using multi-index notation. Since $d{x}^i=d(x^i)$ and hence $d(d{x}^i)=0$, the defining rules forceThis 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, $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$, while the mapping degree of the antipodal map is $-1$. ThereforeBy the stated criterion, ${\pi }^{\ast }\omega =d\eta$. The invariant primitive under a finite group actionstill satisfies $d\bar{\eta }={\pi }^{\ast }\omega$ and descends to a form $\beta$ on ${\mathbb{RP}}^{2n}$. Since pullback through a covering is injective on differential forms, ${\pi }^{\ast }(d\beta -\omega )=0$ implies $d\beta =\omega$. Thus 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 isthe 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 $p^{\ast }\alpha +p^{\ast }\beta \wedge \nu$, andClosedness and exactness are therefore componentwise. Hence the mapis a well-defined bijection $H_{\mathrm{dR}}^k(M)\oplus H_{\mathrm{dR}}^{k-1}(M)\to H_{\mathrm{dR}}^k(M\times S^1)$, proving the de Rham cohomology of a product with a circle formula.