Conformal constraint weights and charged time-symmetric data — Examples

Substituting $q^2=m^2$ into the conformal factor gives ${\psi }^2=(1+m\mathrm{/}(2r){)}^2-m^2\mathrm{/}(4r^2)=1+m\mathrm{/}r$, which is positive for every $r>0$. Therefore $\rho =r{\psi }^2=r+m$ and ${\rho }^{\mathrm{\prime }}(r)=1$.

The radial line element of $h={\psi }^4(d{r}^2+r^2{\gamma }_{S^2})$ is ${\psi }^{2}dr=(1+m\mathrm{/}r)dr$. Since $s^{\mathrm{\prime }}(r)=1+m\mathrm{/}r={\psi }^2>0$, $s$ is a proper radial coordinate; moreover $s\to -\mathrm{\infty }$ as $r\downarrow 0$ and $s\to +\mathrm{\infty }$ as $r\to \mathrm{\infty }$, so its range is $\mathbb{R}$.

Replacing ${\psi }^{4}d{r}^2$ by $d{s}^2$ and using $\rho =r+m$ gives $h=d{s}^2+\rho (s{)}^2{\gamma }_{S^2}$. As $s\to -\mathrm{\infty }$, $r(s)\to 0$, so $\rho (s{)}^2\to m^2$ and the spherical coefficient converges to that of the product cylinder of radius $m$.

The unit normal to $S_r$ is $\nu ={\psi }^{-2}{\mathrm{\partial }}_r$. Its mean curvature is $H={\mathrm{div}}_h\nu =2{\rho }^{\mathrm{\prime }}\mathrm{/}({\psi }^2\rho )=2r\mathrm{/}(r+m{)}^2>0$ for every $r>0$. Thus no centered coordinate sphere is minimal, while the infinite negative range of $s$ proves that the puncture is an infinite-distance cylindrical end.

The seed vector $\bar{E}={\mathrm{\partial }}_x$ has Euclidean divergence ${\bar{D}}_i{\bar{E}}^i={\mathrm{\partial }}_x(1)=0$, so it satisfies the source-free seed Gauss condition. The function $\psi =e^x$ is smooth and positive on all of ${\mathbb{R}}^3$.

If instead one assigns the unscaled vector $E=\bar{E}$, then $\sqrt{\det h}={\psi }^6=e^{6x}$ and the coordinate divergence formula gives $D_i{E}^i=(\sqrt{\det h}{)}^{-1}{\mathrm{\partial }}_i(\sqrt{\det h}{E}^i)=e^{-6x}{\mathrm{\partial }}_x(e^{6x})=6$. Thus the claimed universal preservation fails: this smooth field violates $D_i{E}^i=0$ everywhere.

The correct conformal field is $E={\psi }^{-6}\bar{E}=e^{-6x}{\mathrm{\partial }}_x$, for which step 1.1's seed identity gives $D_i{E}^i=e^{-6x}{\bar{D}}_i{\bar{E}}^i=0$. Hence the counterexample isolates the missing vector weight and uses no singular source at a removed point.