Charged gauge data and SI-natural conventions — Examples

Since $n$ is unit, $P_T(n)=I-n{n}^T$ has kernel $\mathbb{R}n$ and range $n^{\perp }$, of dimension $3-1=2$, by [F1]. For any fixed unit vector $u$ not parallel to $n_0$, $P_T(n)u$ is nonzero on a neighborhood of $n_0$; there define $e_1(n)=P_T(n)u\mathrm{/}\mathrm{\mid }{P}_T(n)u\mathrm{\mid }$ and $e_2(n)=n\times e_1(n)$. By [F4], $e_2$ is orthogonal to both unit vectors and $\mathrm{\mid }{e}_2{\mathrm{\mid }}^2=1-0=1$. It is therefore a continuous unit vector in $n^{\perp }$, orthogonal to $e_1$, so the two vectors form a local frame.

For the fixed-direction fiber in the example, define $T_n(a,v)=v-na$. Since $n\cdot v=a$ and $n\cdot n=1$, its image lies in $n^{\perp }\subset {\mathbb{C}}^3$, and $T_n(0,z)=z$ for every $z\in n^{\perp }$. Writing $v=T_n(a,v)+na$ and $w=T_n(b,w)+nb$ gives $[(a,v),(b,w){]}_{\eta }=\overline{T_n(a,v)}\cdot T_n(b,w)$. Its radical is therefore $\ker T_n=\mathbb{C}(1,n)$: a radical vector pairs with itself to give $\mathrm{\parallel }{T}_n(a,v){\mathrm{\parallel }}^2=0$, and every kernel vector pairs to zero with all others. Thus $[(a,v)]\mapsto T_n(a,v)$ is a positive isometry of $K_{0,n}\mathrm{/}{N}_n$ onto the two-complex-dimensional plane of step 1.1. In contrast, [F2] identifies the global quotient with ${\mathcal{H}}_T$, the space of square-integrable sections of these planes. A global continuous real frame would contradict [F3], so the local construction cannot extend over the whole sphere.

The box $E$ has measure $1\mathrm{/}64$ by [F2], so $ϵ\in L^2({\mathbb{R}}^3;{\mathbb{C}}^4)$. Its spatial components vanish and its time component equals one on $E$, hence $[ϵ,ϵ{]}_{\eta }={\int }_E(-1)d^3\mathbf{k}=-1\mathrm{/}64<0.$

On $E$, $-{ϵ}^0+n\cdot \mathit{ϵ}=-1$, so the subsidiary condition in [F1] fails on a set of measure $1\mathrm{/}64>0$ by [F2]. The quotient in [F1] is formed only from ${\mathcal{K}}_0$; this vector is outside that space and cannot represent a negative-norm physical quotient class.

Choose a nonzero point $z_0$ on the $x_1$ axis. The delta term vanishes away from the origin, and by [F2] the real function ${\mathrm{\partial }}_1^{2}G$ is continuous and strictly positive near $z_0$. Thus the commutator kernel $-i{\mathrm{\partial }}_1^{2}G$ is continuous and nowhere zero on a sufficiently small neighborhood of $z_0$.

Choose nonnegative smooth tests $f,g$ from [F3], supported in small disjoint balls centered at points whose difference is $z_0$, with the difference of their supports contained in that neighborhood and both positive on smaller balls. The delta term in [F1] pairs to zero because the supports are disjoint. The remaining pairing is $-i$ times the integral of the strictly positive real kernel ${\mathrm{\partial }}_1^{2}G$ against $f(x)g(y)$, which is a positive real number before multiplication by $-i$ and hence is nonzero. Therefore the commutator pairing is nonzero for disjoint supports, so its kernel is not supported on the diagonal. The potential is nonlocal; the separate field-strength locality result is unaffected.

Integrating [F1] over $[0,L{]}^3$ gives ${\int }_{[0,L{]}^3}-\mathrm{\Delta }\varphi d^{3}x=-{\sum }_{j=1}^3{\int }_0^L\text{\hspace{0.17em}⁣}{\int }_{[0,L{]}^2}{\mathrm{\partial }}_j^2\varphi d^2{x}_{\widehat{j}}d{x}_j=0,$ because for each $j$ the integral in $x_j$ is the difference of the periodic values of ${\mathrm{\partial }}_j\varphi$ at $L$ and $0$.

The right side of the same integrated equation is ${ϵ}_0^{-1}{\int }_{[0,L{]}^3}{\rho }_0{d}^{3}x={ϵ}_0^{-1}{\rho }_0{L}^3\mathrm{\ne }0,$ contradicting step 1.1. Thus no smooth periodic solution exists for this nonzero total charge; the same integration proves neutrality is necessary for any periodic solution.