Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 3 / b

Because ${}^{\star }{}^{\star }=1$, the exterior algebra of two-forms has the orthogonal decompositionwhere ${}^{\star }H=H$ for a self-dual differential form and ${}^{\star }G=-G$ for an anti-self-dual differential form. The Hodge star is self-adjoint, soIt follows that

For an $SU(n)$ connection, use the positive normand define the Second Chern number byOrthogonality givesTherefore the Euclidean Yang-Mills actionobeys the Yang-Mills instanton Bogomolny boundEquality holds precisely when $F_{-}=0$ or $F_{+}=0$, according to the sign of $k$; these are the self-dual and anti-self-dual Yang-Mills instantons. Other trace and orientation conventions may reverse $k$ but leave the absolute-value bound unchanged.