Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 313 / 3 / Solution

For the curvature two-formuse the normalizationfor the Second Chern form. Graded cyclicity of the trace gives $\mathrm{Tr}(A^{\wedge 4})=0$ andOn the other hand,andThus the coefficient in the Chern-Simons 3-form must beand

For the gauge transformation $A^{\prime }=gA{g}^{-1}-dg{g}^{-1}$, use $d(g^{-1})=-g^{-1}(dg)g^{-1}$ and the Maurer-Cartan equation. Expanding $d{A}^{\prime }+A^{\prime }\wedge A^{\prime }$ makes the terms linear and quadratic in $dg{g}^{-1}$ cancel, leavingInvariance of the matrix trace under conjugation then gives

A Yang-Mills instanton on ${\mathbb{R}}^4$ has finite Euclidean action, smooth curvature in the interior, and $F\to 0$ sufficiently rapidly at infinity. Its connection therefore approaches a pure gauge on the asymptotic three-sphere,up to a decaying correction, for a map $g:S_{\infty }^3\to SU(2)$. By Stokes theorem,Writing $\theta =dg{g}^{-1}$ gives $d\theta =\theta \wedge \theta$ and, for $A=-\theta$,Hence the instanton number as a winding number at infinity isUnder $SU(2)\cong S^3$, this integer is the degree of a map between oriented manifolds $g:S^3\to S^3$. Reversing the trace or orientation convention reverses the displayed sign.