Kerr geometry and hidden symmetry

Since $0<\sqrt{M^2-a^2}\le M$, with equality in the upper bound exactly when $a=0$, one has $0\le r_{-}<r_{+}$ and $r_{-}=0$ exactly when $a=0$. Thus $\mathrm{\Delta }=(r-r_{+})(r-r_{-})>0$ for $r>r_{+}$. Also $\mathrm{\Sigma }\ge r^2>0$, $A:=r^2+a^2>0$, and $q=\gamma (\xi ,\xi )\ge 0$. The global formula in [F1] is a smooth symmetric tensor because its sphere tensors and scalar functions are smooth, its denominators are nonzero, and its finite sums and products are smooth by [F6]. The stated formula for $N$ is smooth: repeated use of [F5] makes $x\mapsto x^{-1}$ smooth on $(0,\mathrm{\infty })$, [F6] composes this with $\mathrm{\Sigma }\mathrm{\Delta }$, and $T,\mathrm{\Phi }$ are smooth by [F1]; hence $N$ is a smooth vector field in the sense of [F7].

On a polar chart put $q={\sin }^2\theta$. The one-forms $e^0=\sqrt{\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }}(dt-aqd\varphi )$, $e^1=\sqrt{\mathrm{\Sigma }\mathrm{/}\mathrm{\Delta }}dr$, $e^2=\sqrt{\mathrm{\Sigma }}d\theta$, and $e^3=\sin \theta {\mathrm{\Sigma }}^{-1\mathrm{/}2}(Ad\varphi -adt)$ are smooth: repeated application of [F5] gives every derivative of the positive powers $x^{1\mathrm{/}2}$ and $x^{-1\mathrm{/}2}$ on $(0,\mathrm{\infty })$, and [F6] gives smoothness after composition with the positive smooth functions $\mathrm{\Delta },\mathrm{\Sigma }$. The determinant of the coefficient matrix of $(e^0,e^3)$ relative to $(dt,d\varphi )$ is $\sqrt{\mathrm{\Delta }}\sin \theta$, which is nonzero when $0<\theta <\pi$. Expanding the global formula of [F1] gives $g=-(e^0{)}^2+(e^1{)}^2+(e^2{)}^2+(e^3{)}^2$; hence $g$ has signature $(-,+,+,+)$ at every point off the axis.

At either axis point ${\xi }^{\mathrm{♭}}=0$, $\mathrm{\Sigma }=A$, and the global expression in [F1] restricts to $g=-(\mathrm{\Delta }\mathrm{/}A)d{t}^2+(A\mathrm{/}\mathrm{\Delta })d{r}^2+A\gamma$ on the tangent directions there. The first coefficient is negative and the radial and sphere terms are positive because $A,\mathrm{\Delta }>0$ and $\gamma$ is positive definite. Thus $g$ is nondegenerate with the same signature at the axes. The global definition in [F1] already gives smoothness there, so this proves the asserted signature on all of ${\mathcal{M}}_{\mathrm{K}}$.

Put $D:=A^2-a^2\mathrm{\Delta }q$. Since $\mathrm{\Delta }=A-2Mr$, $0<\mathrm{\Delta }<A$ and $0\le q\le 1$, one has $D\ge A^2-a^2\mathrm{\Delta }\ge A^2-a^{2}A=A{r}^2>0$. Off the axis, direct contraction of the covariant metric in [F1] with the specified vector $N$ gives $g(N,T)=-1$, $g(N,\mathrm{\Phi })=0$, and $g(N,{\mathrm{\partial }}_r)=g(N,{\mathrm{\partial }}_{\theta })=0$; hence $g(N,\cdot )=-dt$ there. At an axis point $q=0$, $\mathrm{\Phi }=0$, and $\mathrm{\Sigma }=A$, so $N=(A\mathrm{/}\mathrm{\Delta })T$ and the metric of step 2.2 gives the same identity on all tangent directions. Thus everywhere $g(N,N)=-dt(N)=-D\mathrm{/}(\mathrm{\Sigma }\mathrm{\Delta })<0$. Step 1.1 and [F4] now make $({\mathcal{M}}_{\mathrm{K}},g,N)$ a smooth time-oriented Lorentzian four-manifold. Only at this point apply [F2]: nondegeneracy and $g(N,\cdot )=-dt$ identify $N=-{\mathrm{♯}}_g(dt)$; [F3] supplies the smooth inverse metric in its stated scope.

Define $I_K(\lambda ):=K_{ab}(\gamma (\lambda ))v^a(\lambda )v^b(\lambda )$. The product rule in [F3] gives $I_K^{\mathrm{\prime }}=({\mathrm{\nabla }}_{v}K)(v,v)+2K(D_{\lambda }v,v)$.

By [F2], $D_{\lambda }v=0$. Also $v^a{v}^b{v}^c$ is symmetric in its indices, so $({\mathrm{\nabla }}_{v}K)(v,v)=v^a{v}^b{v}^c{\mathrm{\nabla }}_a{K}_{bc}=v^a{v}^b{v}^c{\mathrm{\nabla }}_{(a}{K}_{bc)}=0$ by [F1]. Thus $I_K^{\mathrm{\prime }}=0$ on $I$.

Apply [F4] to $I_K$ on the connected interval $I$. Therefore $K_{ab}{v}^a{v}^b$ is constant along $\gamma$.

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.

Work first on a polar chart with $0<\theta <\pi$ and change angular coordinate to $u=\cos \theta$. Put $q=1-u^2$, $S=r^2+a^2{u}^2$, $D=r^2-2Mr+a^2$, and $A=r^2+a^2$. In coordinate order $(t,\varphi ,r,u)$ the metric is block diagonal between $(t,\varphi )$ and $(r,u)$: $g=G\oplus h$, where $G=S^{-1}\left(\begin{array}{cc}2Mr-S& -2Marq\\ -2Marq& q(A^2-a^{2}Dq)\end{array}\right)$ and $h=\mathrm{diag}(S\mathrm{/}D,S\mathrm{/}q)$. Direct $2\times 2$ inversion gives $G^{-1}=\left(\begin{array}{cc}-(A^2-a^{2}Dq)\mathrm{/}(SD)& -2Mar\mathrm{/}(SD)\\ -2Mar\mathrm{/}(SD)& (D-a^{2}q)\mathrm{/}(SDq)\end{array}\right)$, with $\det G=-Dq$; the inverse base block is $h^{-1}=\mathrm{diag}(D\mathrm{/}S,q\mathrm{/}S)$.

The derivatives needed below are $S_r=2r$, $S_u=2a^{2}u$, $S_{rr}=2$, $S_{ru}=0$, $S_{uu}=2a^2$, $D_r=2(r-M)$, $D_{rr}=2$, $q_u=-2u$, and $q_{uu}=-2$. On this chart $S,D,q>0$, so every denominator in the metric and inverse block is nonzero.

Write Greek indices $\alpha ,\beta ,\gamma$ for $t,\varphi$ and Latin indices $i,j,k,\mathrm{\ell }$ for $r,u$. Since $G,h$ are independent of $t,\varphi$, the complete nonzero Christoffel families from [F3] are ${\mathrm{\Gamma }}^{\alpha }{}_{\beta i}={\mathrm{\Gamma }}^{\alpha }{}_{i\beta }=\frac{1}{2}(G^{-1}{)}^{\alpha \gamma }{\mathrm{\partial }}_i{G}_{\gamma \beta },{\mathrm{\Gamma }}^i{}_{\alpha \beta }=-\frac{1}{2}(h^{-1}{)}^{ij}{\mathrm{\partial }}_j{G}_{\alpha \beta },{\mathrm{\Gamma }}^i{}_{jk}=\frac{1}{2}(h^{-1}{)}^{i\mathrm{\ell }}({\mathrm{\partial }}_j{h}_{\mathrm{\ell }k}+{\mathrm{\partial }}_k{h}_{\mathrm{\ell }j}-{\mathrm{\partial }}_{\mathrm{\ell }}{h}_{jk});$ all families not forced by lower-index symmetry vanish. The six base coefficients, making the last family explicit, are ${\mathrm{\Gamma }}^r{}_{rr}=\frac{r}{S}-\frac{r-M}{D},{\mathrm{\Gamma }}^r{}_{ru}=\frac{a^{2}u}{S},{\mathrm{\Gamma }}^r{}_{uu}=-\frac{Dr}{Sq},{\mathrm{\Gamma }}^u{}_{rr}=-\frac{a^{2}uq}{SD},{\mathrm{\Gamma }}^u{}_{ru}=\frac{r}{S},{\mathrm{\Gamma }}^u{}_{uu}=\frac{a^{2}u}{S}+\frac{u}{q}\mathrm{.}$ Thus the first two matrix families are fully specified by the displayed $G,G^{-1},h^{-1}$ and their $r,u$ derivatives, rather than by omitted connection coefficients.