Explicit charge-one BPST connection on the round four-sphere

The Pauli identity ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ gives $(x^{4}I-i{x}^j{\sigma }_j{)}^{\ast }(x^{4}I-i{x}^j{\sigma }_j)=r^{2}I$. In entries this matrix is $\left(\begin{array}{cc}{x}^4-i{x}^3& -i{x}^1-x^2\\ -i{x}^1+x^2& x^4+i{x}^3\end{array}\right)$, whose determinant is $(x^4{)}^2+(x^3{)}^2+(x^1{)}^2+(x^2{)}^2=r^2$. Thus $u(x)$ is unitary with determinant one wherever $r>0$. The equations $U^{\ast }U=I$ and $\det U=1$ make $SU(2)$ closed in the eight-dimensional real matrix space, and its entries have modulus at most one, so it is bounded; [F2] gives compactness. The traceless skew-Hermitian $2\times 2$ matrices have the form $\left(\begin{array}{cc}ia& z\\ -\bar{z}& -ia\end{array}\right)$ for $a\in \mathbb{R}$ and $z\in \mathbb{C}$, so $T_1,T_2,T_3$ form a real basis. Direct multiplication gives $[T_a,T_b]={ϵ}_{abc}{T}_c$ and $B(T_a,T_b)=\frac{1}{2}{\delta }_{ab}$. For $X,Y\in \mathfrak{s}\mathfrak{u}(2)$, cyclicity gives $B(X,Y)=B(Y,X)$, while $\overline{\mathrm{tr}(XY)}=\mathrm{tr}(Y^{\ast }{X}^{\ast })=\mathrm{tr}(YX)=\mathrm{tr}(XY)$, so $B(X,Y)\in \mathbb{R}$. Since $X^{\ast }=-X$, $B(X,X)=\mathrm{tr}(X^{\ast }X)>0$ for $X\mathrm{\ne }0$; and $B(UX{U}^{-1},UY{U}^{-1})=B(X,Y)$ by cyclicity for $U\in SU(2)$. Thus $B$ is symmetric, positive definite, and adjoint-invariant. Finally $M=\mathrm{diag}(i,-i)$ and $N=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ lie in $SU(2)$, while $MN=-NM$, so $SU(2)$ is non-Abelian. Together with [F1], this verifies the group, representation, and trace hypotheses of the definition.

By [F7], the quotient with $\tau =u^{-1}$ is a smooth right principal $SU(2)$-bundle with sections satisfying $s_S=s_N{u}^{-1}$. Full AC [A1] supplies the countable-choice hypothesis of the gauge-data interface [F6], which specifies the actual global form, singleton defining-representation family, and bundle sector selected in the Statement. Matrix multiplication using ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ gives $u^{-1}du=\frac{2{\eta }_{a\mu \nu }{x}^{\nu }}{r^2}{T}_{a}d{x}^{\mu }$ and $du{u}^{-1}=\frac{2{\bar{\eta }}_{a\mu \nu }{x}^{\nu }}{r^2}{T}_{a}d{x}^{\mu }$. Thus $A_N=\frac{r^2}{r^2+{\rho }^2}{u}^{-1}du$. With transition $g=u^{-1}$, [F3] gives ${\mathrm{Ad}}_u{A}_N+ud(u^{-1})=-\frac{{\rho }^2}{r^2+{\rho }^2}du{u}^{-1}=-\frac{2{\rho }^2{\bar{\eta }}_{a\mu \nu }{x}^{\nu }}{r^2(r^2+{\rho }^2)}{T}_{a}d{x}^{\mu }$. Substituting $y=x\mathrm{/}{r}^2$ and $d{x}^{\mu }=({\delta }_{\mu \lambda }\mathrm{/}\mathrm{\mid }y{\mathrm{\mid }}^2-2y^{\mu }{y}_{\lambda }\mathrm{/}\mathrm{\mid }y{\mathrm{\mid }}^4)d{y}^{\lambda }$, the term containing $y^{\mu }{y}^{\nu }{\bar{\eta }}_{a\mu \nu }$ vanishes by antisymmetry, leaving exactly the displayed $A_S$. The $A_N$ coefficients are smooth at $x=0$ because $\rho >0$, and $A_S$ is smooth at $y=0$ because its denominator is positive and its numerator is linear in $y$. In a product chart with local section $s_i$ and group coordinate $h\in SU(2)$, set ${\omega }_i={\mathrm{Ad}}_{h^{-1}}{A}_i+h^{-1}dh$. The checked transition relation makes these forms agree on the overlap; direct substitution gives ${\omega }_i(Z_P)=Z$ for every fundamental vertical vector $Z_P$ and $R_k^{\ast }{\omega }_i={\mathrm{Ad}}_{k^{-1}}{\omega }_i$ for every fixed $k\in SU(2)$. The converse in [F3] therefore gives a smooth principal connection on the glued bundle.

Differentiating $A_{N,\nu }^a=2{\eta }_{a\nu \lambda }{x}^{\lambda }\mathrm{/}(r^2+{\rho }^2)$ and using [F4] gives $F_{N,\mu \nu }^a=-4{\eta }_{a\mu \nu }\mathrm{/}D+4Q_{a\mu \nu }\mathrm{/}{D}^2$, where $D=r^2+{\rho }^2$ and $Q_{a\mu \nu }=-{\eta }_{a\nu \lambda }{x}^{\lambda }{x}_{\mu }+{\eta }_{a\mu \lambda }{x}^{\lambda }{x}_{\nu }+{ϵ}_{bca}{\eta }_{b\mu \lambda }{\eta }_{c\nu \kappa }{x}^{\lambda }{x}^{\kappa }$. Put $t=x^4$, write $x_a=x^a$ for spatial indices, and set $\mathrm{\mid }\vec{x}{\mathrm{\mid }}^2={\sum }_{k=1}^3(x^k{)}^2$. For $\mu =i,\nu =4$, expanding the symbols gives $Q_{ai4}=x_a{x}_i+t({ϵ}_{aik}{x}_k+{\delta }_{ai}t)-{ϵ}_{bka}({ϵ}_{bil}{x}_l+{\delta }_{bi}t)x_k$. The contraction ${\sum }_b{ϵ}_{bka}{ϵ}_{bil}={\delta }_{ki}{\delta }_{al}-{\delta }_{kl}{\delta }_{ai}$ makes the $x_a{x}_i$ terms cancel, and ${ϵ}_{ika}={ϵ}_{aik}$ cancels the terms linear in $t$; hence $Q_{ai4}=(\mathrm{\mid }\vec{x}{\mathrm{\mid }}^2+t^2){\delta }_{ai}=r^2{\eta }_{ai4}$. For distinct spatial $i,j$, expansion gives the derivative terms $-{ϵ}_{ajl}{x}_l{x}_i-{\delta }_{aj}t{x}_i+{ϵ}_{ail}{x}_l{x}_j+{\delta }_{ai}t{x}_j$. The terms in the bracket product linear in $t$ equal $t(-{\delta }_{ai}{x}_j+{\delta }_{aj}{x}_i)$, so they cancel those derivative terms. Contracting the two spatial epsilon factors by ${\sum }_b{ϵ}_{bca}{ϵ}_{bil}={\delta }_{ci}{\delta }_{al}-{\delta }_{cl}{\delta }_{ai}$ leaves $R_{aij}={ϵ}_{ijm}{x}_a{x}_m-{ϵ}_{ajl}{x}_l{x}_i+{ϵ}_{ail}{x}_l{x}_j$, and the two-index-$4$ term is $t^2{ϵ}_{aij}$. Let $k$ be the remaining spatial index and $\epsilon ={ϵ}_{ijk}$. If $a=i$, the first two terms of $R$ are $\epsilon x_i{x}_k-\epsilon x_k{x}_i=0$; if $a=j$, they are $\epsilon x_j{x}_k-\epsilon x_k{x}_j=0$; if $a=k$, all three terms sum to $\epsilon (x_i^2+x_j^2+x_k^2)$. Thus $R_{aij}={ϵ}_{aij}\mathrm{\mid }\vec{x}{\mathrm{\mid }}^2$ and $Q_{aij}=r^2{\eta }_{aij}$. These are all independent antisymmetric index pairs, so $F_{N,\mu \nu }^a=-4{\rho }^2{\eta }_{a\mu \nu }\mathrm{/}(r^2+{\rho }^2{)}^2$. The two-forms ${\eta }_1=d{x}^{14}+d{x}^{23}$, ${\eta }_2=d{x}^{24}+d{x}^{31}$, and ${\eta }_3=d{x}^{34}+d{x}^{12}$ are self-dual in the chosen Euclidean orientation because the Euclidean Hodge star interchanges the two displayed summands in each. Write $g_X=\mathrm{\Omega }(x{)}^2\delta$ on the north chart, where $\mathrm{\Omega }(x)=2\mathrm{/}(1+r^2)>0$ and $\delta ={\sum }_{\mu }(d{x}^{\mu }{)}^2$. At each point, the induced inner product on two-forms scales by ${\mathrm{\Omega }}^{-4}$ while the oriented volume form scales by ${\mathrm{\Omega }}^4$. Thus $⟨\alpha ,\beta {⟩}_{g_X}d{\mathrm{vol}}_{g_X}=⟨\alpha ,\beta {⟩}_{\delta }{d}^{4}x$ for all two-forms $\alpha ,\beta$ there. The defining wedge identity and nondegeneracy of the wedge pairing imply ${\ast }_{g_X}={\ast }_{\delta }$ on two-forms pointwise. Hence $F_N={\ast }_{g_X}{F}_N$.

The trace form gives $B(T_a,T_b)=\frac{1}{2}{\delta }_{ab}$ and the defining symbols satisfy ${\sum }_{a,\mu ,\nu }{\eta }_{a\mu \nu }^2=12$. Since $F_N=\frac{1}{2}{F}_{N,\mu \nu }d{x}^{\mu }\wedge d{x}^{\nu }$, the local action density is $B(F_N\wedge {\ast }_{g_X}{F}_N)=48{\rho }^4(r^2+{\rho }^2{)}^{-4}{d}^{4}x$. The radial substitution $t=r^2\mathrm{/}{\rho }^2$ gives ${\int }_{{\mathbb{R}}^4}{\rho }^4(r^2+{\rho }^2{)}^{-4}{d}^{4}x=2{\pi }^2\cdot \frac{1}{2}{\int }_0^{\mathrm{\infty }}t(1+t{)}^{-4}dt={\pi }^2\mathrm{/}6$: with $v=1+t$, ${\int }_0^{\mathrm{\infty }}t(1+t{)}^{-4}dt={\int }_1^{\mathrm{\infty }}(v^{-3}-v^{-4})dv=[-1\mathrm{/}(2v^2)+1\mathrm{/}(3v^3){]}_1^{\mathrm{\infty }}=1\mathrm{/}2-1\mathrm{/}3=1\mathrm{/}6$. Hence ${\int }_{S^4}B(F\wedge {\ast }_{g_X}F)=8{\pi }^2$; the omitted stereographic point has measure zero and the patched curvature is smooth. Self-duality gives $B(F\wedge F)=B(F\wedge {\ast }_{g_X}F)$, so $q_B=1$. The action is $S_B=8{\pi }^2\mathrm{/}{g}^2$, finite and equal to the bound in [F5]. The definition and bound interfaces invoked in [F5] both assume full AC [A1].