A form is <harmonic differential form>[harmonic] when
$$
\Delta\alpha=(d\delta+\delta d)\alpha=0.
$$
Adjointness gives
$$
\langle\Delta\alpha,\alpha\rangle_X
=\lVert d\alpha\rVert^2+\lVert\delta\alpha\rVert^2.
$$
Therefore $\Delta\alpha=0$ implies $d\alpha=0$ and $\delta\alpha=0$; the converse follows immediately from the definition of $\Delta$.

The <Hodge decomposition theorem> states in particular that every de Rham cohomology class has a unique harmonic representative. Hence
$$
\mathcal H^p(X)\xrightarrow{\sim}H^p_{\mathrm{dR}}(X),
\qquad \alpha\longmapsto[\alpha].
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-115/4/d/solution