Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 118 / 1 / b

The assertion requires $p+q>0$. As printed, it also includes $(p,q)=(0,0)$, where the nonzero constant function $1$ is closed but cannot be the exterior derivative of a negative-degree form. The zero constant is harmless; below we prove the intended positive-degree result.

Set $k=p+q>0$, let ${\rho }_t(z)=tz$, and letThe polydisc is preserved by these dilations. The radial homotopy operator isHere ${\iota }_R$ is the interior product of a differential form. For a smooth $k$-form, its pulled-back contraction is $O(t^k)$, so the integral and its coefficient derivatives converge at zero. The identity for a pullback of a differential form along the radial flow and Cartan's magic formula giveIntegrating yields $dH\alpha +Hd\alpha =\alpha -{\rho }_0^{\ast }\alpha$. The final pullback vanishes in positive degree; since $d\alpha =0$, we have $dH\alpha =\alpha$.

Each ${\rho }_t$ is a holomorphic map, so its pullback of a differential form preserves bidegree. Contracting with the $(1,0)$ part of $R$ lowers $p$ by one, and contracting with its $(0,1)$ part lowers $q$ by one. Therefore the primitive has precisely the permitted types:This is a type-preserving refinement of the Poincaré lemma; when one index is zero, the corresponding negative-degree summand is simply absent.