= Solution
The curvature of $D=d+A$ is
$$
\boxed{F=D^2=dA+A\wedge A},
$$
or $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu]$. The covariant exterior derivative and the <Jacobi identity> give the <Bianchi identity>
$$
\boxed{D_AF=dF+A\wedge F-F\wedge A=0},
\qquad D_{[\mu}F_{\nu\rho]}=0.
$$
Since the <Hodge star operator> satisfies $*^2=1$ on Euclidean two-forms,
$$
0\leq\int_{\mathbb R^4}-\operatorname{Tr}[(F\mp*F)\wedge*(F\mp*F)]
=S_E\mp16\pi^2k.
$$
Using the opposite choice of sign as well gives
$$
\boxed{S_E\geq16\pi^2|k|}.
$$
Equality holds precisely for a self-dual or anti-self-dual <Yang-Mills instanton>, with the sign selected by that of $k$.
Now assume $\partial_0A_\mu=0$ and write $\Phi=A_0$. Then
$$
F_{0i}=-D_i\Phi,
\qquad
B_i=\frac12\epsilon_{ijk}F_{jk}.
$$
The spatial Bianchi identity and the self-duality equations become
$$
\boxed{D_iB_i=0,
\qquad B_i=\mp D_i\Phi},
$$
where the upper four-dimensional sign gives the first displayed reduced sign under the orientation used here. Consequently
$$
\boxed{\sum_{i=1}^3D_i^2\Phi=0}.
$$
Gauge invariance of the inner product gives
$$
\Delta|\Phi|^2
=2\sum_i|D_i\Phi|^2
+2\left\langle\Phi,\sum_iD_i^2\Phi\right\rangle
=2\sum_i|D_i\Phi|^2.
$$
For $w=1-|\Phi|^2$,
$$
\boxed{\Delta w=-2\sum_i|D_i\Phi|^2\leq0}.
$$
Since $w\to0$ at infinity, the <weak maximum principle for elliptic operators> excludes a negative interior minimum. Hence
$$
\boxed{|\Phi|\leq1\quad\text{everywhere}}.
$$
With $d^4x=dx^0d^3x$, direct decomposition gives
$$
\boxed{-2\operatorname{Tr}(F\wedge*F)
=(|D\Phi|^2+|B|^2)d^4x},
$$
$$
\boxed{\operatorname{Tr}(F\wedge F)
=\langle D_i\Phi,B_i\rangle d^4x}.
$$
Because $D_iB_i=0$,
$$
\langle D_i\Phi,B_i\rangle
=\partial_i\langle\Phi,B_i\rangle,
$$
so the Pontryagin density reduces to the surface charge $4\pi N$ per unit $x^0$. The conventional three-dimensional energy
$$
E_3=\frac12\int_{\mathbb R^3}(|D\Phi|^2+|B|^2)d^3x
$$
obeys the <Bogomolny bound> $E_3\geq4\pi|N|$, 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 $[T_a,T_b]=-\epsilon_{abc}T_c$ gives
$$
\boxed{B_i^a=(2\alpha+r\alpha')\delta_{ia}
-\left(\frac{\alpha'}r+\alpha^2\right)x_ix_a}.
$$
On a sphere of radius $r$,
$$
B_i^an_i=(2\alpha-r^2\alpha^2)n_a,
\qquad
\Phi^a=f(r)n_a.
$$
Therefore
$$
N=\lim_{r\to\infty}\frac1{4\pi}
\int_{S_r^2}f(r)(2\alpha-r^2\alpha^2)dS.
$$
The boundary conditions $f\to1$ and $r^2\alpha\to1$ make the integrand equal to $r^{-2}+o(r^{-2})$. Thus
$$
\boxed{N=1}.
$$
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