Kerr horizon angular velocity and the ergoregion

The coefficients in [F1] have no $t$ or $\varphi$ dependence. For $T={\mathrm{\partial }}_t$ and $\mathrm{\Phi }={\mathrm{\partial }}_{\varphi }$, the coordinate components are constant, so every term in [F4] vanishes; hence ${\mathcal{L}}_{T}g={\mathcal{L}}_{\mathrm{\Phi }}g=0$ off the axis. These Lie derivatives are smooth tensor fields because $g,T,\mathrm{\Phi }$ are smooth globally. The axis-avoiding set is dense, so continuity extends their vanishing to the axis; both fields are therefore Killing everywhere by [F3].

Choose $0<\epsilon <r_{+}\mathrm{/}2$ and use the extension manifold $\bar{\mathcal{M}}:={\mathbb{R}}_v\times (r_{+}-\epsilon ,\mathrm{\infty })\times S^2$ with its product smooth structure. Let $\bar{\pi }:\bar{\mathcal{M}}\to S^2$ be the sphere projection. For $y=(v,r,p)$, define the product tangent isomorphism ${\bar{I}}_y:T_y\bar{\mathcal{M}}\to \mathbb{R}\oplus \mathbb{R}\oplus T_p{S}^2$ by ${\bar{I}}_y(w):=(d{v}_y(w),d{r}_y(w),d{\bar{\pi }}_y(w))$. Set ${\bar{T}}_y:={\bar{I}}_y^{-1}(1,0,0)$ and ${\bar{\mathrm{\Phi }}}_y:={\bar{I}}_y^{-1}(0,0,{\xi }_p)$; these are the product lifts of the stationary and axial fields. Let $\bar{\gamma }:={\bar{\pi }}^{\ast }{\gamma }_{S^2}$, $\eta :={\bar{\pi }}^{\ast }({\gamma }_{S^2}(\xi ,\cdot ))$, and $q:=\bar{\gamma }(\bar{\mathrm{\Phi }},\bar{\mathrm{\Phi }})=\mathrm{\mid }\xi {\mathrm{\mid }}_{{\gamma }_{S^2}}^2$. These are global smooth tensors or functions; on any local polar chart of the new sphere factor, $\eta ={\sin }^2\theta d\tilde{\varphi }$ and $q={\sin }^2\theta$. Fix $r_0>r_{+}$ and define $f(r)={\int }_{r_0}^r(s^2+a^2)\mathrm{/}\mathrm{\Delta }(s)ds$ and $h(r)={\int }_{r_0}^{r}a\mathrm{/}\mathrm{\Delta }(s)ds$ on $(r_{+},\mathrm{\infty })$. Their integrands are smooth because $\mathrm{\Delta }>0$: repeated use of [F7] gives all derivatives of ${\mathrm{\Delta }}^{-1}$, and [F8] handles the polynomial products. [F6] gives $f,h\in C^1$, and successive differentiation of $f^{\mathrm{\prime }}=(r^2+a^2)\mathrm{/}\mathrm{\Delta }$ and $h^{\mathrm{\prime }}=a\mathrm{/}\mathrm{\Delta }$ shows that both primitives are smooth. Set $v=t+f(r)$ and, on the sphere factor, replace each point $(x,y,z)$ by $R_{h(r)}(x,y,z):=(x\cos h-y\sin h,x\sin h+y\cos h,z)$. Repeated application of [F10] makes this rotation smoothly dependent on $r$, and its inverse is $R_{-h(r)}$. Since $R_{h(r)}$ is a rotation about the same axis as $\xi$, its differential preserves $\xi$; thus the induced exterior diffeomorphism carries $\mathrm{\Phi }$ to $\bar{\mathrm{\Phi }}$. It also carries $T$ to $\bar{T}$. Hence the exterior map is a smooth diffeomorphism, with coordinate Jacobian determinant $1$ on every local polar chart; there $\tilde{\varphi }=\varphi +h(r)$ and the pullback of the original ${\xi }^{\mathrm{♭}}$ is $\eta -aq{\mathrm{\Delta }}^{-1}dr$. Substitution in [F1], using ${\mathrm{\Sigma }}^2+a^{2}q(\mathrm{\Sigma }+2Mr)=(r^2+a^2{)}^2-a^2\mathrm{\Delta }q$, gives the global ingoing expression $g=-\frac{\mathrm{\Delta }-a^{2}q}{\mathrm{\Sigma }}dv\otimes dv+dv\otimes dr+dr\otimes dv+\mathrm{\Sigma }\bar{\gamma }+\frac{a^2(\mathrm{\Sigma }+2Mr)}{\mathrm{\Sigma }}\eta \otimes \eta -\frac{2aMr}{\mathrm{\Sigma }}(dv\otimes \eta +\eta \otimes dv)-a(dr\otimes \eta +\eta \otimes dr)\mathrm{.}$ For every $r>r_{+}-\epsilon >r_{+}\mathrm{/}2>0$, one has $\mathrm{\Sigma }>0$ and $\mathrm{\Delta }=A-2Mr<A$, where $A:=r^2+a^2$. If $\mathrm{\Delta }\ge 0$, then $A^2-a^2\mathrm{\Delta }q\ge A^2-a^2\mathrm{\Delta }\ge A^2-a^{2}A=A{r}^2>0$; if $\mathrm{\Delta }<0$, then $A^2-a^2\mathrm{\Delta }q\ge A^2>0$. Consequently the angular restriction in an off-axis polar chart, $\mathrm{\Sigma }d{\theta }^2+\frac{q(A^2-a^2\mathrm{\Delta }q)}{\mathrm{\Sigma }}d{\tilde{\varphi }}^2$, is positive definite. The determinant of the full metric matrix there is $-{\mathrm{\Sigma }}^{2}q<0$; its Schur complement in the $(v,r)$ directions is symmetric with negative determinant, hence has one negative and one positive eigenvalue. The full metric therefore has signature $(-,+,+,+)$ off the axis. At an axis point $\eta =0$ and $\mathrm{\Sigma }=A$; the $(v,r)$ block has matrix $\left(\begin{array}{cc}-\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }& 1\\ 1& 0\end{array}\right)$ of determinant $-1$, and the sphere block is $\mathrm{\Sigma }{\gamma }_{S^2}$ on the tangent space, so the extension has the same Lorentzian signature there. Every coefficient in the global expression is smooth across $r=r_{+}$ because $\mathrm{\Sigma }>0$ and no coefficient contains $\mathrm{\Delta }$ in a denominator. It is a smooth Lorentzian extension across $r=r_{+}$. Write $R:={\mathrm{\partial }}_r$ in the ingoing product coordinates and set $\bar{N}:=\bar{T}-(2Mr\mathrm{/}\mathrm{\Sigma })R$. The displayed covariant metric gives $g(\bar{T},R)=1$, $g(R,R)=0$, and $g(\bar{T},\bar{T})=-(\mathrm{\Delta }-a^{2}q)\mathrm{/}\mathrm{\Sigma }$. Since $\mathrm{\Sigma }=r^2+a^2(1-q)$, direct contraction gives $g(\bar{N},\bar{N})=-\frac{\mathrm{\Delta }-a^{2}q+4Mr}{\mathrm{\Sigma }}=-\frac{\mathrm{\Sigma }+2Mr}{\mathrm{\Sigma }}<0$ on the entire extension, where $r>r_{+}-\epsilon >0$. Both product coordinate fields and $\mathrm{\Sigma }$ are smooth globally, including the rotation axis, so this is a smooth nowhere-zero timelike field there. Thus the extension with $\bar{N}$ is a time-oriented smooth Lorentzian four-manifold, and [F9] now applies. In this chart $g^{-1}(dr,dr)=\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }$, so $dr({\mathrm{♯}}_{g}dr)=g^{-1}(dr,dr)=0$ on $H^{+}$ by [F9]; the normal is null and tangent there.

Put $A=r^2+a^2$. In the ingoing chart the inverse metric has $g^{vr}=A\mathrm{/}\mathrm{\Sigma }$, $g^{\tilde{\varphi }r}=a\mathrm{/}\mathrm{\Sigma }$, and $g^{rr}=\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }$. Thus on $H^{+}$, ${\mathrm{♯}}_g(dr)=(A\mathrm{/}\mathrm{\Sigma }){\mathrm{\partial }}_v+(a\mathrm{/}\mathrm{\Sigma }){\mathrm{\partial }}_{\tilde{\varphi }}$ by [F9]. By step 2.1, $\bar{T}={\mathrm{\partial }}_v$ and $\bar{\mathrm{\Phi }}={\mathrm{\partial }}_{\tilde{\varphi }}$, so ${\mathrm{♯}}_g(dr)=(A\mathrm{/}\mathrm{\Sigma })(\bar{T}+(a\mathrm{/}A)\bar{\mathrm{\Phi }})$. Hence the null normal direction is $\bar{\chi }:=\bar{T}+{\mathrm{\Omega }}_H\bar{\mathrm{\Phi }}$, where ${\mathrm{\Omega }}_H=a\mathrm{/}(r_{+}^2+a^2)$; on the exterior overlap $\bar{\chi }=\chi =T+{\mathrm{\Omega }}_H\mathrm{\Phi }$. Equivalently, $\mathrm{\Delta }(r_{+})=0$ gives $2M{r}_{+}=r_{+}^2+a^2$, and direct substitution yields $g(\bar{\chi },\bar{\chi }){\mathrm{\mid }}_{r_{+}}=0$. It remains to check that this time orientation agrees with the exterior one. On the exterior overlap, $t=v-f(r)$ and $f^{\mathrm{\prime }}=(r^2+a^2)\mathrm{/}\mathrm{\Delta }$ from step 2.1 imply $dt(\bar{N})=1+\frac{2Mr(r^2+a^2)}{\mathrm{\Sigma }\mathrm{\Delta }}>0.$ Thus $g(N,\bar{N})=-dt(\bar{N})<0$ by [F2], so the two timelike fields select the same future cone on the overlap. Finally put $A_{+}:=r_{+}^2+a^2$ and ${\mathrm{\Sigma }}_{+}:=r_{+}^2+a^2(1-q)$. On $H^{+}$, $g(\bar{\chi },R)=1-a{\mathrm{\Omega }}_{H}q=\frac{{\mathrm{\Sigma }}_{+}}{A_{+}}>0.$ Since $\bar{\chi }$ is normal to $H^{+}$ and $\bar{T}$ is tangent, $g(\bar{\chi },\bar{T})=0$ there; using $2M{r}_{+}=A_{+}$ gives $g(\bar{\chi },\bar{N})=-\frac{A_{+}}{{\mathrm{\Sigma }}_{+}}\frac{{\mathrm{\Sigma }}_{+}}{A_{+}}=-1<0.$ Therefore the nonzero null generator $\bar{\chi }$ is future-directed in this proved compatible time orientation; it agrees with $\chi$ on the exterior overlap and requires no extra orientation hypothesis.

From [F1], $g(T,T)=-(1-2Mr\mathrm{/}\mathrm{\Sigma })$. It is positive precisely when $r^2-2Mr+a^2{\cos }^2\theta <0$. The two roots of this quadratic in $r$ are $M\pm \sqrt{M^2-a^2{\cos }^2\theta }$; because $\mathrm{\mid }a\mathrm{\mid }<M$, they are real, and the upper root is $r_{\mathrm{s}}(\theta )$. The lower root is at most $r_{-}$, since ${\cos }^2\theta \le 1$, and $r_{-}<r_{+}$; restricting the interval between the roots to the exterior therefore gives exactly $r_{+}<r<r_{\mathrm{s}}(\theta )$. At $\theta =0,\pi$ the upper root equals $r_{+}$. If $a\mathrm{\ne }0$ and $0<\theta <\pi$, then $a^2{\cos }^2\theta <a^2$, so the upper root exceeds $r_{+}$; if $a=0$, the upper root equals $r_{+}$ for every $\theta$ and the ergoregion is empty. This proves the stated condition in all allowed cases.