Pole identities and the Ernst scalar inequality reduction

At $p$, $T=0$ gives $f=0$ and $A_0=\mathrm{id}$. Since $\sigma =f+i\psi$ with real $\psi$, $\bar{\sigma }=-\sigma$ there. The metric formula gives $g_0=(d_0\mathrm{/}4)g$, and denominator positivity gives $d_0>0$. Substitution into the inverse relation yields the displayed pole relation; in particular $1-\sigma$ and $1+\sigma =1-\bar{\sigma }$ are nonzero.

Evaluate the real part of the pole relation on $(n,\nu )$. By the normal form its right-hand side has real value $\mathrm{Re}((1-\sigma )\mathrm{/}(1+\sigma ))$. The real part of $\mathcal{F}(n,\nu )$ is $({\mathrm{\nabla }}_n{T}^{\mathrm{♭}})(\nu )=g({\mathrm{\nabla }}_{n}T,\nu )$. Because $\sigma$ is purely imaginary at $p$, $1\mathrm{/}(1-\sigma )$ and $1\mathrm{/}(1+\sigma )$ are conjugates, so the displayed real part equals $2\mathrm{Re}(1\mathrm{/}(1-\sigma ))-1$. The difference of two connections applied to $T$ is tensorial and vanishes where $T=0$; hence ${\mathrm{\nabla }}_n^{(0)}T={\mathrm{\nabla }}_{n}T$ at $p$, and the metric scaling gives the stated contraction identity.

The flows of $K$ and $Z$ preserve $g_0$: they preserve every factor in the Harrison metric formula. Let ${\varphi }_t$ be the flow of $Z$, fixing $p$. Since $K=T+{\mathrm{\Omega }}_{H}Z$ and this flow preserves $T$, $[Z,K]=0$. Thus its differential $P_t=(d{\varphi }_t{)}_p$ commutes with the differential of the $K$-flow at $p$, and hence with its infinitesimal generator $B(v)={\mathrm{\nabla }}_{v}K{\mathrm{\mid }}_p$. Since $K{\mathrm{\mid }}_S=0$, $B$ vanishes on $T_{p}S$; the Killing equation makes $B$ skew with respect to $g$, so it preserves the $g$-normal plane. Its restriction there is the supplied nonzero boost, which is invertible, giving $\ker B=T_{p}S$. Commutation therefore makes $P_t$ preserve $T_{p}S$ and, because $P_t$ is a $g$-isometry, its $g$-orthogonal normal plane. The positive conformal relation at $p$ makes this the $g_0$-normal plane as well. The continuous family $P_t$, starting at the identity, preserves its future cone. At the fixed point $p$, average any future $g_0$-unit timelike normal over this periodic differential action. In an orthonormal normal frame, each averaged vector has time component larger than the absolute value of its spatial component; compactness and the triangle inequality make the average future timelike. Normalize it to obtain an invariant future unit normal $\bar{n}$, and let $\bar{\nu }$ be its outward unit orthogonal normal. Differentiating the fixed vector relation under the flow gives ${\mathrm{\nabla }}_{\bar{n}}^{(0)}Z=0$ at $p$, since $Z(p)=0$.

At $p$, the connection difference also vanishes on $K$, because $K(p)=0$. Thus the supplied boost action holds for ${\mathrm{\nabla }}^{(0)}K$ as well. The two metrics are positive constant multiples at this point, so their future/outward normal planes agree and the same boost rate sends $\bar{n}$ to $\kappa \bar{\nu }$. Using $K=T+{\mathrm{\Omega }}_{H}Z$ and step 2.2 gives $g_0({\mathrm{\nabla }}_{\bar{n}}^{(0)}T,\bar{\nu })=\kappa >0$. Step 2.1 therefore gives the positive value $4\kappa \mathrm{/}{d}_0$ for the scalar expression. Since that expression is independent of the chosen oriented normal frame, the same positivity holds for the original $n,\nu$, proving the inequality.