A causal barrier identifies the future Kerr event horizon

Fix a point of $H$ and a sufficiently small product coordinate box with compact closure and with $r_{+}$ in its radial interval. Put $N=-\mathrm{\nabla }\tau$, $A=\sqrt{-g(N,N)}>0$ and $E_0=N\mathrm{/}A$, a smooth future unit timelike field by [F1]. By [F6], $E_0^{\perp }$ is three-dimensional and positive definite. A smooth orthonormal spatial frame can be constructed locally as follows. Project the four coordinate vector fields $Z_a$ to $P_a=Z_a+g(Z_a,E_0)E_0$; these span $E_0^{\perp }$. Select three independent projections at the chosen point and shrink the box so they remain independent. Relabel them $P_1,P_2,P_3$ and successively set $W_j=P_j-{\sum }_{1\le i<j}g(P_j,E_i)E_i$ and $E_j=W_j\mathrm{/}\sqrt{g(W_j,W_j)}$. Positive definiteness makes each denominator positive; the resulting fields are smooth and orthonormal. Writing $V=t{E}_0+{\sum }_i{x}_i{E}_i$ gives $g(V,V)=-t^2+\mathrm{\mid }x{\mathrm{\mid }}^2$ and $g(V,N)=-At$. Thus [F6] says that the nonzero future causal vectors are exactly those with $t>0$ and $\mathrm{\mid }x\mathrm{\mid }\le t$, and $d\tau (V)=-g(N,V)=At>0$. Consequently the vectors normalized by $d\tau (X)=1$ are exactly $X=A^{-1}(E_0+{\sum }_i{u}_i{E}_i)$ with ${\sum }_i{u}_i^2\le 1$. For two future causal vectors $U=t{E}_0+{\sum }_i{x}_i{E}_i$ and $V=s{E}_0+{\sum }_i{z}_i{E}_i$, [F7] gives $g(U,V)=-ts+x\cdot z\le -ts+\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }\le 0$; the same inequality holds if either vector is zero. This establishes the required cone representation and pairing directly under the supplied conventions.

By the normalized representation in step 1.1 and [F7], the maximum radial component is $m=A^{-1}(dr(E_0)+\sqrt{{\sum }_{i}dr(E_i{)}^2})$: for $b=(dr(E_i){)}_{i=1}^3$, the maximum of $b\cdot u$ on $\mathrm{\mid }u\mathrm{\mid }\le 1$ is $\mathrm{\mid }b\mathrm{\mid }$, attained at $u=b\mathrm{/}\mathrm{\mid }b\mathrm{\mid }$ when $b\mathrm{\ne }0$, and at $u=0$ when $b=0$. This function is locally Lipschitz: its coefficients are smooth and the Euclidean norm is Lipschitz by [F7]; bounded first derivatives on a smaller compact coordinate box and the mean value theorem [F4] give a common local bound. On $H$, the future null normal in [F1] and the pairing inequality of step 1.1 give $dr(X)=g(\mathrm{\nabla }r,X)\le 0$. Equality is attained by $X=\mathrm{\nabla }r\mathrm{/}d\tau (\mathrm{\nabla }r)$ since this is a nonzero future null vector and $dr(\mathrm{\nabla }r)=g(\mathrm{\nabla }r,\mathrm{\nabla }r)=0$ there. Thus $m=0$ on $H$. After shrinking the box, comparison with the point having the same other coordinates and $r=r_{+}$ gives $m\le C\mathrm{\mid }r-r_{+}\mathrm{\mid }$ for a fixed $C\ge 0$. Scaling a nonzero $V$ by $d\tau (V)$, and treating $V=0$ separately, yields $dr(V)\le C\mathrm{\mid }r-r_{+}\mathrm{\mid }d\tau (V)$. Thus along every compact causal segment in this box, $q^{\mathrm{\prime }}\ge 0$ and $y^{\mathrm{\prime }}\le C\mathrm{\mid }y\mathrm{\mid }{q}^{\mathrm{\prime }}$ almost everywhere. No change of curve parameter or regularity of an inverse parameter map is used.

Put $w(s)=\max \{y(s),0\}$. The positive-part map is $1$-Lipschitz, so [F2] makes $w$ absolutely continuous. Outside a null set, $y,w,q$ are differentiable and the inequalities of step 2.1 hold. At such a point with $y>0$ one has $w^{\mathrm{\prime }}=y^{\mathrm{\prime }}\le Cw{q}^{\mathrm{\prime }}$; with $y<0$ one has $w^{\mathrm{\prime }}=0$. At a point where $y=0$, differentiability of $w$ and its minimum value $0$ imply $w^{\mathrm{\prime }}=0$, again giving the same inequality. Therefore $w^{\mathrm{\prime }}\le Cw{q}^{\mathrm{\prime }}$ almost everywhere, including the zero level. On any fixed compact interval $[s_0,s_1]$ in the box, $q$ has compact range. By [F4], the function $t\mapsto e^{-C(t-q(s_0))}$ is Lipschitz on that range; [F2] therefore makes its composition with $q$ absolutely continuous, and also makes $z(s)=e^{-C(q(s)-q(s_0))}w(s)$ absolutely continuous. The pointwise rules [F4] now give $z^{\mathrm{\prime }}=e^{-C(q-q(s_0))}(w^{\mathrm{\prime }}-C{q}^{\mathrm{\prime }}w)\le 0$ almost everywhere. By the fundamental theorem with the choice hypotheses [F3] and the integral convention [F5], $z(s)-z(s_0)={\int }_{s_0}^s{z}^{\mathrm{\prime }}(u)du\le 0$ for every $s\in [s_0,s_1]$. If $y(s_0)=0$, then $z(s_0)=0$ and $z\ge 0$, so $w=0$ on the interval. This proves the local outward barrier for the originally parametrized absolutely continuous curve, including its constant pieces.

Suppose a future causal curve has a point with $r\le r_{+}$ and a later point with $r>r_{+}$. Restrict to the compact parameter interval $[a,b]$ between these points, so $y(a)\le 0<y(b)$. By continuity and [F8], choose $0<\eta <b-a$ so that $y>0$ on $[b-\eta ,b]$. The set $S=\{s\in [a,b]:y(s)\le 0\}$ is nonempty because $a\in S$, and it is bounded above by $b-\eta$. Thus [F8] gives $s_{\ast }=\sup S$ with $a\le s_{\ast }\le b-\eta <b$. If $y(s_{\ast })>0$, continuity supplies $\delta >0$ such that $y(s)>0$ for $s\in [a,b]$ with $\mathrm{\mid }s-s_{\ast }\mathrm{\mid }<\delta$; the supremum approximation in [F8] supplies $s\in S$ with $s_{\ast }-\delta <s\le s_{\ast }$, a contradiction. Hence $y(s_{\ast })\le 0$. If $y(s_{\ast })<0$, continuity gives some $h$ with $0<h<b-s_{\ast }$ and $y(s_{\ast }+h)<0$, contradicting that $s_{\ast }$ is an upper bound of $S$. Therefore $y(s_{\ast })=0$. Every $s\in (s_{\ast },b]$ has $y(s)>0$, since otherwise $s\in S$ would exceed its upper bound. Continuity of the curve puts $[s_{\ast },s_{\ast }+\epsilon ]$, for some $0<\epsilon <b-s_{\ast }$, in a coordinate box of step 1.1 about $\gamma (s_{\ast })$. Step 3.1 forces $y\le 0$ there, contradicting the positivity just proved. Thus no future causal curve starting at or below $H$ reaches the exterior. This supremum argument does not assume that a first strictly exterior point exists.

Every path ending at the chosen conformal infinity eventually lies in its collar, where $r$ is arbitrarily large. Hence a point with $r\le r_{+}$ cannot reach that infinity by step 4.1. Every exterior point does reach it by the outgoing null curves of [F1]. Taking the complement gives $\{0<r\le r_{+}\}$, whose boundary in the product manifold ${\mathcal{K}}^{\mathrm{i}\mathrm{n}}$ is $H=\{r=r_{+}\}$. This proves the event-horizon identification for $\mathrm{\mid }a\mathrm{\mid }<M$ and $\mathrm{\mid }a\mathrm{\mid }=M$ without assuming a bifurcate horizon or identifying different infinities in a maximal extension.