The outer Kerr null hypersurface and its future generator

Multiplication by the metric matrix yields $n:=\mathrm{\nabla }r=((r^2+a^2){\mathrm{\partial }}_v+\mathrm{\Delta }{\mathrm{\partial }}_r+a\xi )\mathrm{/}\mathrm{\Sigma }$. This pole-free expression and $g(n,n)=\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }$ hold also on the axis by smoothness. Since $dr$ never vanishes and $\mathrm{\Delta }(r_{+})=0$, the product level $H=\{r=r_{+}\}$ is smooth and null, and $n{\mathrm{\mid }}_H=b\chi$ with $b=(r_{+}^2+a^2)\mathrm{/}\mathrm{\Sigma }>0$. Also $dr(n)=0$ there, so $n$ is tangent and normal to $H$. In a Lorentzian tangent space, the orthogonal complement of a nonzero null vector has that vector as its radical and has positive definite quotient by its span; consequently all nonzero null tangents to $H$ lie on the line spanned by $n$. This identifies the null tangent line without yet asserting its geodesic property.

With $\tau =v-r$ one has $d\tau (\chi )=1>0$. Since $\tau$ increases on future causal curves in the chosen orientation, $\chi$ is future null. Its $v$ component is one, so it is nonzero even at the rotation axis, where $\xi$ vanishes.

For smooth local fields $X,Y$, metric compatibility and torsion freedom give $g({\mathrm{\nabla }}_{X}n,Y)-g({\mathrm{\nabla }}_{Y}n,X)=X(Yr)-Y(Xr)-[X,Y]r=0$. Thus $g({\mathrm{\nabla }}_{n}n,X)=g({\mathrm{\nabla }}_{X}n,n)=\frac{1}{2}X(g(n,n))$ for every $X$, and nondegeneracy yields ${\mathrm{\nabla }}_{n}n=\frac{1}{2}\mathrm{\nabla }(\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma })$. On $H$, this becomes ${\mathrm{\nabla }}_{n}n={\mathrm{\Delta }}^{\mathrm{\prime }}(r_{+})n\mathrm{/}(2\mathrm{\Sigma })$, where ${\mathrm{\Delta }}^{\mathrm{\prime }}(r_{+})=2r_{+}-2M$. Since $n=b\chi$ on $H$, differentiation in the tangent direction $n$ is legitimate for this restricted equality; moreover $\chi (b)=0$ because $b$ depends only on $r$ and the rotation-invariant sphere coordinate $z$. The connection product rule therefore gives ${\mathrm{\nabla }}_{\chi }\chi =\kappa \chi$ on $H$, with $\kappa ={\mathrm{\Delta }}^{\mathrm{\prime }}(r_{+})\mathrm{/}(2(r_{+}^2+a^2))=(r_{+}-r_{-})\mathrm{/}(2(r_{+}^2+a^2))\ge 0$. Thus these null curves are pregeodesic, meaning their acceleration is proportional to their velocity.

Through any $(v_0,r_{+},p_0)\in H$, the curve $c(u)=(v_0+u,r_{+},R_{{\mathrm{\Omega }}_{H}u}{p}_0)$, $u\in \mathbb{R}$, has velocity $\chi$ and is future null. If $\kappa >0$, set $s=(e^{\kappa u}-1)\mathrm{/}\kappa$; if $\kappa =0$, set $s=u$. In both cases $ds\mathrm{/}du>0$, and writing $w=du\mathrm{/}ds$ gives $dc\mathrm{/}ds=w\chi$ and $D_s(dc\mathrm{/}ds)=(w^2\kappa +wdw\mathrm{/}du)\chi =0$ by the preceding connection product rule. Thus this is an affine null geodesic in the sense of [F2]. Its affine parameter interval is $(-1\mathrm{/}\kappa ,\mathrm{\infty })$ when $\kappa >0$ and $\mathbb{R}$ when $\kappa =0$. The only finite boundary in the first case corresponds to $u\to -\mathrm{\infty }$ and hence $v\to -\mathrm{\infty }$, so no continuous extension at that boundary exists in the specified product manifold. These geodesics are therefore maximally continued in $H$. Conversely, any future null curve in $H$ is locally a positive reparametrization of one of these curves: its tangent lies in the line from step 1.1, and taking $u=v-v_0$ gives $dr\mathrm{/}du=0$ and the rotation equation $dp\mathrm{/}du={\mathrm{\Omega }}_H\xi (p)$, whose explicit solution is $p=R_{{\mathrm{\Omega }}_{H}u}{p}_0$ (as is checked by differentiating $R_{-{\mathrm{\Omega }}_{H}u}p$). Any proper segment extends along that same orbit, after matching its affine parameter by a positive affine change of the displayed $s$. Hence the complete orbit images are exactly the generators of [F2], and $\chi$ is tangent to them. All formulas remain valid for $\mathrm{\mid }a\mathrm{\mid }=M$, when $\kappa =0$; none divides by $r_{+}-r_{-}$.