 Solution

The curvature of $D=d+A$ isor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$. The covariant exterior derivative and the Jacobi identity give the Bianchi identity

Since the Hodge star operator satisfies ${\ast }^2=1$ on Euclidean two-forms,Using the opposite choice of sign as well givesEquality holds precisely for a self-dual or anti-self-dual Yang-Mills instanton, with the sign selected by that of $k$.

Now assume ${\partial }_0{A}_{\mu }=0$ and write $\Phi =A_0$. ThenThe spatial Bianchi identity and the self-duality equations becomewhere the upper four-dimensional sign gives the first displayed reduced sign under the orientation used here. Consequently

Gauge invariance of the inner product givesFor $w=1-|\Phi {|}^2$,Since $w\to 0$ at infinity, the weak maximum principle for elliptic operators excludes a negative interior minimum. Hence

With $d^{4}x=d{x}^0{d}^{3}x$, direct decomposition givesBecause $D_i{B}_i=0$,so the Pontryagin density reduces to the surface charge $4\pi N$ per unit $x^0$. The conventional three-dimensional energyobeys the Bogomolny bound $E_3\ge 4\pi |N|$, saturated by the Bogomolny-Prasad-Sommerfield monopole equation. The four-dimensional action density per unit $x^0$ is $2E_3$; signs relating $k$ and $N$ depend on the self-duality and orientation convention.

For the hedgehog ansatz for a monopole in the question, direct differentiation with $[T_a,T_b]=-{ϵ}_{abc}{T}_c$ givesOn a sphere of radius $r$,ThereforeThe boundary conditions $f\to 1$ and $r^2\alpha \to 1$ make the integrand equal to $r^{-2}+o(r^{-2})$. Thus