1 Introduction

The Kerr–Newman–de Sitter (KNdS) metric is a solution of the Einstein–Maxwell equations with a positive cosmological constant \(\Lambda\):

$$\begin{aligned}{} & {} R_{\mu \nu }-\,\frac{1}{2}Rg_{\mu \nu }+\Lambda g_{\mu \nu }=2T_{\mu \nu },\\{} & {} dF=d\star F=0,\\{} & {} T_{\mu \nu }=F_{\mu \alpha }F_\nu ^{\ \alpha }-\,\frac{1}{4}F_{\alpha \beta } F^{\alpha \beta }g_{\mu \nu }. \end{aligned}$$

Here \(R_{\mu \nu }\) are the components of the Ricci tensor of the spacetime metric g, R is the scalar curvature, \(\star\) is the Hodge star operator, and \(F_{\mu \nu }\) is the Faraday electromagnetic 2-form. So, the KNdS metric is an electrovacuum solution of the Einstein field equations, i.e. it is a solution in which the only nongravitational mass-energy present is an electromagnetic field. The spacetime is the four-dimensional manifold \({{\mathbb {R}}}^2\times {{\mathbb {S}}}^2\) with metric given by

$$\begin{aligned} g& = \frac{\rho ^2}{\Delta _r}{\,\textrm{d}}r^2+\frac{\rho ^2}{\Delta _{\theta }}{\,\textrm{d}}\theta ^2 +\frac{1}{\rho ^2}(a^2\sin ^2\theta \Delta _{\theta }-\Delta _r){\,\textrm{d}}t^2\nonumber \\{} & {} +\frac{\sin ^2\theta }{\Xi ^2\rho ^2}((r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _r){\,\textrm{d}}\phi ^2 -\,\frac{2a\sin ^2\theta }{\Xi \rho ^2}((r^2+a^2)\Delta _{\theta }-\Delta _r){\,\textrm{d}}\phi {\,\textrm{d}}t \end{aligned}$$
(1)

in Boyer–Lindquist coordinates, where

$$\begin{aligned} \Delta _r& = (r^2+a^2)\left( 1-\,\frac{\Lambda }{3}r^2\right) -2Mr+e^2,\\ \Delta _{\theta }& = 1+\frac{\Lambda }{3}a^2\cos ^2\theta ,\\ \rho ^2& = r^2+a^2\cos ^2\theta ,\\ \Xi& = 1+\frac{\Lambda }{3}a^2, \end{aligned}$$

\(\theta\) is the colatitude, with

$$\begin{aligned} 0\le \theta \le \pi , \end{aligned}$$

and \(\phi\) is the longitude, with

$$\begin{aligned} \phi \in {{\mathbb {S}}}^1, \end{aligned}$$

(see Carter [6], and Akcay and Matzner [1] and Kraniotis [23], for example). Here \(M>0\) and \(a\ne 0\) are mass and angular momentum parameters, respectively, and e is a charge parameter (which may be zero). Without loss of generality, we assume that the magnetic charge is zero and a is positive. This metric is supposed to represent a rotating black hole, with charge, in a universe which is expanding at an accelerated rate (as ours is). We refer to "Appendix C", where we calculate Komar integrals over the event horizon, for the relation between these parameters and the physical quantities of the black hole. We wish to consider subextremal metrics, meaning that \(\Delta _r\) has four distinct real roots

$$\begin{aligned} r_n<0<r_-<r_+<r_c. \end{aligned}$$

The event horizon \(\mathcal{H}\) corresponds to the hypersurface where \(r=r_+\), the Cauchy horizon \(\mathcal{C}\mathcal{H}\) corresponds to the hypersurface where \(r=r_-\), and the cosmological horizon corresponds to the hypersurface \(r=r_c\). We are interested in studying solutions of the wave equation in the black hole region, \(r_-<r<r_+\), and we wish to construct double null coordinates and understand their relation with Boyer–Lindquist coordinates. The coordinate t takes values in \(\mathbb {R}\).

This article follows closely the strategy and tools developed in [9]. Double null coordinate systems were constructed by Pretorius and Israel [28] for Kerr spacetimes, by Balushi and Mann [3] for Kerr–(anti) de Sitter spacetimes, and by Imseis, Balushi and Mann [19] for Kerr–Newman–(anti) de Sitter spacetimes. In [3] and [19] the authors also study the formation of caustics. In Sect. 2 we construct a double null coordinate system for Kerr–Newman–de Sitter spacetimes. This construction only differs from the one in [3] and [19] (that we were unaware of until the completion of this work) in the choice of \(\lambda\) (our choice \(\lambda =\sin ^2\theta _\star\) is identical to the one in [9] and [28]). Consider the transformation \((t,r,\theta ,\phi )\mapsto (t,r_\star ,\theta _\star ,\phi )\) in the black hole region \(r_-<r<r_+\), where \(\Delta _r<0\). The coordinate \(\theta _{\star }\) is defined implicitly as the solution of \(F(r,\theta ,\theta _{\star })=0\), where F is given by

$$\begin{aligned} F(r, \theta ,\theta _{\star }) =\int _{\theta _{\star }}^{\theta } \frac{{\,\textrm{d}}\theta '}{a\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}} +\int _r^{r_+}\frac{{\,\textrm{d}}r'}{\sqrt{((r')^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _{r'}}}. \end{aligned}$$

The coordinate \(r_\star\) is defined by \(r_\star =\varrho (r,\theta ,\sin ^2\theta _{\star }(r,\theta ))\), where \(\varrho\) is given by

$$\begin{aligned} \varrho (r,\theta ,\lambda )& = \int ^r_{r_0}\frac{(r^{\prime 2}+a^2)}{\Delta _{r'}}{\,\textrm{d}}r'+ \int _r^{r_+}\frac{(r^{\prime 2}+a^2)- \sqrt{((r')^2+a^2)^2-a^2\lambda \Delta _{r'}}}{\Delta _{r'}}{\,\textrm{d}}r'\\{} & {} +\int _0^\theta \frac{a\sqrt{\lambda \Delta _{\theta '}-\sin ^2\theta '}}{\Delta _{\theta '}}{\,\textrm{d}}\theta '+\frac{a^2}{2}f(\lambda ), \end{aligned}$$

for some fixed \(r_0\in (r_-,r_+)\), and where f is the function which satisfies \(f(0)=0\) and

$$\begin{aligned} f'(\lambda )=- \int _0^{\arcsin \sqrt{\lambda }} \frac{a\sqrt{\lambda \Delta _{\theta '}-\sin ^2\theta '}}{\Delta _{\theta '}}{{\,\textrm{d}}\theta '} \end{aligned}$$

(note the difference, \(\varrho\) versus \(\rho\) in (1)).

The regularity of transformation of coordinates \((r_\star ,\theta _\star )\mapsto (r,\theta )\) for Kerr spacetimes, namely at \(\theta _\star =0\), was shown by Dafermos and Luk [9]. We adapt their work to the setting of Kerr–Newman–de Sitter spacetimes. We check that \(\theta _{\star }\) is well defined and continuous at \(\theta =0\) and \(\theta =\frac{\pi }{2}\) with

$$\begin{aligned} \frac{\sin \theta _{\star }}{\sin \theta }+\frac{\cos \theta }{\cos \theta _{\star }} \lesssim 1. \end{aligned}$$
(2)

The proof of (2) requires that we use conditions characterizing subextremal black holes which are deduced in "Appendix  B". Namely, we use

$$\begin{aligned} \Xi <1+\frac{1}{3}\,l\left( \frac{r_+}{r_-},\Lambda e^2\right) , \end{aligned}$$
(3)

where the function l is given by (138). This implies the inequality

$$\begin{aligned} \Xi <\csc ^2\left( \arctan \sqrt{\frac{r_+}{r_-}}+\arctan \sqrt{\frac{r_-}{r_+}}\right) , \end{aligned}$$

which in turn implies (2). We use the fact that \(\Lambda\) is nonnegative so that our computations are not immediately applicable to the setting of Kerr–AdS.

We also show that r and \(\theta\) are smooth functions of \(r_\star\) and \(\theta _{\star }\). When the cosmological constant is equal to zero and \(e=0\), \(\Delta _{\theta }\) is equal to 1 and our formulas reduce to the ones for the Kerr spacetime in [9]. The trigonometric identity

$$\begin{aligned} \sin ^2(2 \theta _{\star }) \Delta _\theta =-\sin (2\theta )\,\partial _\theta D(\theta ,\theta _{\star })+ 2 (\cos (2 \theta )+\cos (2 \theta _{\star }))D(\theta ,\theta _{\star }), \end{aligned}$$

for \(D(\theta ,\theta _{\star })=\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta\) is the key to the calculation of \(\partial _r\theta _\star\) and \(\partial _\theta \theta _\star\), as well as the successful completion of some new identities, such as (54) and (59), which we need in order to calculate the derivatives of r and \(\theta\) with respect to \(r_\star\) and \(\theta _\star\). We would like to emphasize that our calculations are successful because the dependence of \(\Delta _{\theta }\) on \(\theta\) occurs through \(\sin ^2\theta\), and not through \(\sin \theta\), or on any other non-smooth function of \(\sin ^2\theta\). (This is a reflection of the fact that the Kerr–Newman–de Sitter metric is regular on a manifold diffeomorphic to \(\mathbb {R}^2\times \mathbb {S}^2_{\theta ,\phi }\), i.e. it is regular on the full Boyer–Lindquist spheres of constant time coordinate t and radial coordinate r.) We also obtain bounds on the derivatives of r and \(\theta\) which we need later on. These bounds parallel the ones in [9].

Using

$$\begin{aligned} u=\frac{r_\star -t}{2}\qquad \hbox {and}\qquad v=\frac{t+r_\star }{2}, \end{aligned}$$

the final transformation

$$\begin{aligned} \phi _\star =\phi -h(r_\star ,\phi _\star ), \end{aligned}$$

with h given by

$$\begin{aligned} \partial _{r_{\star }} h(r_{\star },\theta _{\star })=-\,\frac{\Xi a ((r^2+a^2) \Delta _{\theta }-\Delta _r)}{(r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}},\qquad \hbox {with}\ h(0,\theta _{\star })=0, \end{aligned}$$

allows one to bring the metric to the double null form

$$\begin{aligned} g& = -2\Omega ^2 \left( {\textrm{d}}u\otimes {\textrm{d}}v+{\,\textrm{d}}v\otimes {\textrm{d}}u\right) \nonumber \\{} & {} +{g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} {\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\theta _{\star } +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}\theta _{\star }\otimes ({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v) +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v)\otimes {\textrm{d}}\theta _{\star }\nonumber \\{} & {} +{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }} ({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v)\otimes ({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v), \end{aligned}$$
(4)

which one can use to carry out energy estimates.

Neither the Boyer–Lindquist coordinate system \((t,r,\theta ,\phi )\) nor the double null coordinate system \((u,v,\theta _\star ,\phi _\star )\) cover the axis \(\theta _\star =0\), obviously. But they can be naturally extended to an atlas that does cover \(\theta _\star =0\) using

$$\begin{aligned} \left\{ \begin{array}{rcl} x&{}=&{}\sin \theta _\star \cos \phi ,\\ y&{}=&{}\sin \theta _\star \sin \phi , \end{array}\right. \qquad \qquad \left\{ \begin{array}{rcl} \tilde{x}&{}=&{}\sin \theta \cos \phi ,\\ \tilde{y}&{}=&{}\sin \theta \sin \phi . \end{array}\right. \end{aligned}$$

In Sect. 2.2.6, we prove that the two atlases \(\mathcal{A}_{\tiny \hbox {BL}}=\{(t,r,\theta ,\phi ),(t,r,\tilde{x},\tilde{y})\}\) and \(\mathcal{A}_{\tiny \hbox { DN}}=\{(u,v,\theta _\star ,\phi _\star ),(u,v,x,y)\}\) are compatible (which would be clear if we were to exclude the points where \(\theta =\theta _\star =0\) and \(\theta =\theta _\star =\pi\) from our manifold). This implies that the two-spheres given by the intersection of the hypersurfaces \(u=\hbox {constant}\) and \(v=\hbox {constant}\) are \(C^\infty\) with respect to \(\mathcal{A}_{\tiny \hbox { BL}}\) (see Theorem 2.23).

Following [9], we analyze the decay of \(\Omega ^2\) at the future event and Cauchy horizons, \(\mathcal{H}^+\) and \(\mathcal{C}\mathcal{H}^+\). Finally, we give regular coordinates at \(\mathcal{H}^+\) and \(\mathcal{C}\mathcal{H}^+\). The Christoffel symbols of g in the double null coordinates \((u,v,\theta _\star ,\phi _\star )\) are given in "Appendix A", along with some covariant derivatives that are needed to carry out energy estimates.

In Sect. 3, using the vector field method, we study, in the black hole interior, the energy of solutions of the wave equation which have compact support on \(\mathcal{H}^+\). We apply the form (4) of the metric to construct certain blue-shift and red-shift vector fields and to calculate their covariant derivatives. We obtain the usual inequalities relating the vector and scalar currents associated to these vector fields. This allows us to illustrate how Sbierski’s result in [29] applies to Kerr–Newman–de Sitter spacetimes.

This work is a first step of our broader project to generalize the results of [7], which provides a sufficient condition, in terms of surface gravities and a parameter for an exponential decaying Price Law, for energy of waves to remain bounded up to \(\mathcal{C}\mathcal{H}^+\). The work [7] used the fact that the generators of spherical symmetry are three Killing vector fields, which is not true in the context of Kerr–Newman–de Sitter spacetimes. We expect to address this in a forthcoming paper.

An alternative approach towards extending the results of [7] to the Kerr–Newman–de Sitter setting would be to work in Boyer–Lindquist coordinates as is done in the work [25] on Kerr black hole interiors. However, as an example, the double null coordinates allow one to immediately extend the results in [14] to KNdS.

In "Appendix  B", we characterize subextremal Kerr–Newman–de Sitter black holes in terms of \((r_-, r_+,\Lambda a^2,\Lambda e^2)\), proving (3), in particular, as mentioned above. The subset of \({{\mathbb {R}}}^3\) where one can choose \(\left( \frac{r_+}{r_-},\Lambda a^2,\Lambda e^2\right)\) is sketched in Fig. 2 on page 43. We make additional remarks concerning alternative choices of parameters, namely \(\left( \Lambda ,\frac{r_+}{r_-},a,e\right)\) or \((\Lambda ,M,a,e)\). Related characterizations of the parameters of subextremal Kerr–de Sitter solutions, for the case when there is no charge, can be found in Lake and Zannias [24] and Borthwick [4].

Hintz and Vasy give a uniform analysis of linear waves up to the Cauchy horizon using methods from scattering theory and microlocal analysis in [18]. Moreover, Hintz proves non-linear stability of the Kerr–Newman–de Sitter family of charged black holes in [17].

2 A double null coordinate system

2.1 Construction of the double null coordinate system

2.1.1 The coordinates \(r_\star\) and \(\theta _{\star }\)

Given the manifold \(\mathbb {R}^2\times \mathbb {S}^2\), with metric (1), we look for a function \(r_\star\) such that the axisymmetric hypersurface,

$$\begin{aligned} v(t,r,\theta )=t\pm r_\star (r,\theta )=\hbox {constant} \end{aligned}$$

(ingoing when the plus sign is chosen, and outgoing when when the minus sign is chosen), is lightlike. Then, the function v must satisfy the eikonal equation

$$\begin{aligned} g^{\alpha \beta }(\partial _{\alpha } v)(\partial _{\beta } v) =\frac{1}{\rho ^2}\left[ \Delta _r (\partial _r {r}_{\star })^2+\Delta _{\theta }(\partial _{\theta } {r}_{\star })^2 -\,\frac{1}{\Delta _{r}\Delta _{\theta }}\left( (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}\right) \right] =0. \end{aligned}$$

We follow [28] and construct particular separable solutions of the eikonal equation. We define P and Q by

$$\begin{aligned} P(\theta ,\theta _{\star })& = a\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }, \\ Q(r,\theta _{\star })& = \sqrt{(r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r}. \end{aligned}$$

Note that

$$\begin{aligned} \frac{1}{\Delta _{r}\Delta _{\theta }}\left( (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}\right) =\frac{Q^2}{\Delta _{r}}+\frac{P^2}{\Delta _{\theta }}, \end{aligned}$$
(5)

and so the eikonal equation becomes

$$\begin{aligned} \Delta _r (\partial _r {r}_{\star })^2+\Delta _{\theta }(\partial _{\theta } {r}_{\star })^2=\frac{Q^2}{\Delta _{r}}+\frac{P^2}{\Delta _{\theta }}. \end{aligned}$$

As P is independent of r, and Q is independent of \(\theta\), we look for special solutions \(r_\star\) of this equation, where

$$\begin{aligned} \partial _r {r}_{\star }=\frac{Q}{\Delta _r}\quad \hbox {and}\quad \partial _{\theta } {r}_{\star }=\frac{P}{\Delta _\theta }, \end{aligned}$$
(6)

so that

$$\begin{aligned} {\,\textrm{d}}r_{\star }=\frac{Q}{\Delta _r}{\,\textrm{d}}r+\frac{P}{\Delta _\theta }{\,\textrm{d}}\theta . \end{aligned}$$
(7)

Both P and Q depend on (what is so far the parameter) \(\theta _\star\), which arises because of the degree of freedom one has in breaking up the left-hand side of (5) to a sum. Indeed, to the left-hand side of (5) we subtracted and added the quantity \(a^2\sin ^2\theta _{\star }\) (which is independent of both r and \(\theta\)) and then we decomposed the resulting expression into a sum of a function depending solely on r and a function depending solely on \(\theta\). We integrate (7) and obtain

$$\begin{aligned} r_\star =\int _{r_+}^r\frac{Q(r',\theta _{\star })}{\Delta _{r'}}{\,\textrm{d}}r'+\int _0^\theta \frac{P(\theta ',\theta _{\star })}{\Delta _{\theta '}}{\,\textrm{d}}\theta '+ \frac{a^2}{2}f(\sin ^2\theta _{\star }), \end{aligned}$$

where the function f accounts for an integration constant. Thus we have

$$\begin{aligned} r_\star =\varrho (r,\theta ,\lambda ), \end{aligned}$$
(8)

\(\varrho :(r_-,r_+)\times [0,\pi ]\times [0,1]\rightarrow \mathbb {R}\), with

$$\begin{aligned} \varrho (r,\theta ,\lambda )& = \int ^r_{r_0}\frac{(r^{\prime 2}+a^2)}{\Delta _{r'}}{\,\textrm{d}}r'+ \int _r^{r_+}\frac{(r^{\prime 2}+a^2)-\hat{Q}(r',\lambda )}{\Delta _{r'}}{\,\textrm{d}}r'\\{} & {} +\int _0^\theta \frac{\hat{P}(\theta ',\lambda )}{\Delta _{\theta '}}{\,\textrm{d}}\theta '+\frac{a^2}{2}f(\lambda ), \end{aligned}$$

for some fixed \(r_0\in (r_-,r_+)\), and

$$\begin{aligned} \lambda =\sin ^2\theta _{\star }. \end{aligned}$$

The expression for \(\varrho\) is written so that the second integral converges. Here

$$\begin{aligned} \hat{P}^2(\theta ,\lambda )& = a^2\left( \lambda \Delta _{\theta }-\sin ^2\theta \right) , \\ \hat{Q}^2(r,\lambda )& = \left( (r^2+a^2)^2-a^2\lambda \Delta _r\right) , \end{aligned}$$

so that \(\hat{P}(\theta ,\lambda )= P\bigl (\theta ,\arcsin \sqrt{\lambda }\bigr ) =P(\theta ,\theta _{\star })\) and \(\hat{Q}(r,\lambda )= Q\bigl (r,\arcsin \sqrt{\lambda }\bigr )=Q(r,\theta _{\star })\).

For each fixed \(\lambda\), (8) is a solution of (7). We now proceed to obtain another solution of (7). Calculating the differential of \(\varrho\), we obtain

$$\begin{aligned} {\,\textrm{d}}\varrho =\frac{Q}{\Delta _r}{\,\textrm{d}}r+\frac{P}{\Delta _\theta }{\,\textrm{d}}\theta + \partial _\lambda \varrho {\,\textrm{d}}\lambda , \end{aligned}$$

where

$$\begin{aligned} \partial _\lambda \varrho =\frac{a^2}{2}\left( \int _0^\theta \frac{{\,\textrm{d}}\theta '}{\hat{P}(\theta ',\lambda )}+ \int _r^{r_+}\frac{{\,\textrm{d}}r'}{\hat{Q}(r',\lambda )}+f'(\lambda ) \right) . \end{aligned}$$

Define the function

$$\begin{aligned} F(r,\theta ,\theta _\star ):= \int _0^\theta \frac{{\,\textrm{d}}\theta '}{P(\theta ',\theta _{\star })}+ \int _r^{r_+}\frac{{\,\textrm{d}}r'}{Q(r',\theta _{\star })}+f'(\sin ^2\theta _{\star }). \end{aligned}$$

Note that the function f is still free. We choose \(f:[0,1]\rightarrow \mathbb {R}\) to be the function which satisfies \(f(0)=0\) and

$$\begin{aligned} f'(\lambda )=- \int _0^{\arcsin \sqrt{\lambda }} \frac{{\,\textrm{d}}\theta '}{\hat{P}\bigl (\theta ', \lambda )},\qquad \hbox {i.e.}\ f'(\sin ^2\theta _{\star })=-\int _{0}^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}. \end{aligned}$$

The function f is bounded. Then the expression for F becomes

$$\begin{aligned} F(r, \theta ,\theta _{\star })& = \int _{\theta _{\star }}^{\theta } \frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}+\int _r^{r_+}\frac{{\,\textrm{d}}r'}{Q(r', \theta _{\star })}. \end{aligned}$$
(9)

Adapting the construction of [9] to our case, for \(r\in [r_-,r_+]\) and \(\theta \in (0, \frac{\pi }{2})\), we define \(\theta _{\star }\in [\theta , \frac{\pi }{2})\) implicitly to be the solution of

$$\begin{aligned} F(r, \theta ,\theta _{\star })=0 \end{aligned}$$
(10)

(see Lemma 2.4). Also, let

$$\begin{aligned} \theta _{\star }(r,0)=0,\quad \theta _{\star }\left( r,\frac{\pi }{2}\right) = \frac{\pi }{2} \end{aligned}$$

and

$$\begin{aligned} \theta _{\star }(r,\theta )=\pi -\theta _{\star }(r,\pi -\theta )\ \hbox {for}\ \theta \in \left( \frac{\pi }{2},\pi \right] . \end{aligned}$$
(11)

Then, the function

$$\begin{aligned} r_\star =\varrho (r,\theta ,\sin ^2\theta _{\star }(r,\theta )) \end{aligned}$$

is another solution of (7). The functions

$$\begin{aligned} \qquad u=\frac{r_{\star }(r,\theta )-t}{2}\quad \hbox {and}\quad v=\frac{t+{r}_{\star }(r, \theta )}{2} \end{aligned}$$
(12)

are solutions of the eikonal equation. Just as in the case of Kerr, it turns out that \(\theta _{\star }\) is an appropriate angle coordinate. This can be understood starting with the construction of [28]: when \(\Lambda =M=a=e=0\) (so that we are reduced to the Minkowski spacetime), \(\theta _{\star }\) is the spherical polar angle. Moreover, for r close to \(r_+\), \(\theta _{\star }\) is close to \(\theta\) (see (9) and (10)). The function \(\theta _\star\) is interpreted as the spherical polar angle and the hypersurfaces where u and v are constant are called quasi-spherical light cones. From

$$\begin{aligned} g^{\alpha \beta }(\partial _{\alpha } r)(\partial _{\beta } r)& = \frac{1}{\rho ^2} (\Delta _r (\partial _r {r}_{\star })^2+\Delta _{\theta }(\partial _{\theta } {r}_{\star })^2)\ =\ \frac{1}{\rho ^2\Delta _r\Delta _{\theta }}(\Delta _rP^2+\Delta _{\theta }Q^2)\\& = \frac{1}{\rho ^2\Delta _{r}\Delta _{\theta }}\left( (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}\right) \ <\ 0 \end{aligned}$$

(recall that \(\Delta _r<0\)), a hypersurface where \(r_{\star }\) equals a constant is spacelike.

Remark 2.1

Note that \(r_{\star }\) ranges between \(-\infty\) and \(+\infty\), as r ranges between \(r_+\) and \(r_-\). More precisely, given \(L>0\), there exists \(\delta >0\) such that \(r_\star (r,\theta )>L\) for all \((r,\theta )\in (r_-,r_-+\delta )\times [0,\pi ]\). Moreover, given \(\delta >0\), there exists \(L>0\) such that \(r_\star (r,\theta )>L\) implies that \(r\in (r_-,r_-+\delta )\). In Lemma 2.9 we will prove that \((r,\theta )\mapsto (r_\star ,\theta _\star )\) is invertible. So, we are observing that \(\lim _{r\searrow r_-}r_\star (r,\theta )=+\infty\) and that \(\lim _{r_\star \nearrow +\infty }r(r_\star ,\theta _\star )=r_-\), and that these limits are uniform in \(\theta\) and in \(\theta _\star\), respectively, for \(\theta\) and \(\theta _\star\) in \([0,\pi ]\). Analogous statements can be made for the other endpoint.

The behavior of the coordinates u and v is sketched in Fig. 1.

Fig. 1
figure 1

Behavior of the coordinates u and v

Remark 2.2

Because of the symmetry in (11), our statements about \(\theta\) and \(\theta _\star\) will refer to the interval \(\left( 0,\frac{\pi }{2}\right)\) and it is understood that corresponding statements will hold in \(\left( \frac{\pi }{2},\pi \right)\).

2.1.2 The metric in \((t,r_\star ,\theta _{\star },\phi )\) coordinates

Denote

$$\begin{aligned} \Upsilon := \Delta _{r}P^2+\Delta _{\theta }Q^2= (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}, \end{aligned}$$
(13)
$$\begin{aligned} G:=\partial _{\theta _{\star }} F \end{aligned}$$

and

$$\begin{aligned} L:= -GPQ. \end{aligned}$$
(14)

Differentiating both sides of (9) with respect to r and \(\theta\) yields

$$\begin{aligned} \partial _r\theta _\star = \frac{1}{GQ},\qquad \partial _\theta \theta _\star =-\,\frac{1}{GP}. \end{aligned}$$
(15)

Since

$$\begin{aligned} \left| \begin{array}{ll} \partial _rr_\star &{}\partial _\theta r_\star \\ \partial _r\theta _\star &{}\partial _\theta \theta _\star \end{array} \right| = \frac{\Upsilon }{\Delta _r\Delta _{\theta }L}, \end{aligned}$$

the differentials of r and \(\theta\) are given by

$$\begin{aligned} {\,\textrm{d}}r& = \frac{\Delta _{r}\Delta _{\theta }Q}{\Upsilon }{\,\textrm{d}}r_\star -\,\frac{\Delta _{r}LP}{\Upsilon }{\,\textrm{d}}\theta _{\star }, \end{aligned}$$
(16)
$$\begin{aligned} {\,\textrm{d}}\theta& = \frac{\Delta _{r}\Delta _{\theta }P}{\Upsilon }{\,\textrm{d}}r_\star +\frac{\Delta _{\theta }LQ}{\Upsilon }{\,\textrm{d}}\theta _{\star }. \end{aligned}$$
(17)

To write the metric in \((t,r_{\star },\theta _{\star },\phi )\) coordinates one uses (16) and (17) in (1) and obtains

$$\begin{aligned} g& = \frac{\rho ^2\Delta _r\Delta _{\theta }}{\Upsilon }{\,\textrm{d}}r_{\star }^2+\frac{\rho ^2L^2}{\Upsilon }{\,\textrm{d}}\theta _{\star }^2 +\frac{1}{\rho ^2}(a^2\sin ^2\theta \Delta _{\theta }-\Delta _r){\,\textrm{d}}t^2 +\frac{\sin ^2\theta }{\Xi ^2\rho ^2}\Upsilon {\,\textrm{d}}\phi ^2\\{} & {} -\,\frac{2a\sin ^2\theta }{\Xi \rho ^2}((r^2+a^2)\Delta _{\theta }-\Delta _r){\,\textrm{d}}\phi {\,\textrm{d}}t\\& = g_{r_{\star } r_{\star }} \left( {\textrm{d}}r_{\star }\otimes {\textrm{d}}r_{\star }-{\,\textrm{d}}t\otimes {\textrm{d}}t\right) +g_{\theta _{\star } \theta _{\star }}{\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\theta _{\star } +g_{\phi \phi }({\textrm{d}}\phi -B{\,\textrm{d}}t)\otimes ({\textrm{d}}\phi -B{\,\textrm{d}}t), \end{aligned}$$

with

$$\begin{aligned} B=-\,\frac{g_{\phi t}}{g_{\phi \phi }}=\frac{\Xi a((r^2+a^2) \Delta _{\theta }-\Delta _r)}{\Upsilon }. \end{aligned}$$

2.1.3 Definition of \(\phi _{\star }\) and the metric in double null coordinates \((u,v,\theta _{\star },\phi _{\star })\)

From (12), one gets

$$\begin{aligned} \left\{ \begin{array}{lcl} dr_{\star }&{}=&{}{\,\textrm{d}}u+{\,\textrm{d}}v,\\ dt&{}=&{}{\,\textrm{d}}v-{\,\textrm{d}}u. \end{array}\right. \end{aligned}$$

Now introduce a new coordinate \(\phi _{\star }\), defined by

$$\begin{aligned} \phi _{\star }=\phi -h(r_{\star },\theta _{\star }), \end{aligned}$$

where

$$\begin{aligned} \partial _{r_{\star }} h(r_{\star },\theta _{\star })=-B=-\,\frac{\Xi a ((r^2+a^2) \Delta _{\theta }-\Delta _r)}{\Upsilon },\qquad \hbox {with}\ h(0,\theta _{\star })=0. \end{aligned}$$
(18)

For a general function f, one has

$$\begin{aligned} \hat{f}(\hat{u},\hat{v},\hat{\theta }_\star ,\phi )= \hat{f}(\hat{u},\hat{v},\hat{\theta }_\star ,\phi _\star +h(\hat{u}+\hat{v},\hat{\theta }_\star ))= f(u,v,\theta _\star ,\phi _\star ), \end{aligned}$$

where \(\hat{f}\) is the function f written in the \((\hat{u},\hat{v},\hat{\theta }_\star ,\phi )\) coordinate system and \(\hat{u}=u\), \(\hat{v}=v\), \(\hat{\theta }_\star =\theta _\star\). So, it follows that

$$\begin{aligned} \partial _{u}& = \partial _{\hat{u}}+\partial _{r_\star }h\,\partial _\phi ,\\ \partial _{v}& = \partial _{\hat{v}}+\partial _{r_\star }h\,\partial _\phi ,\\ \partial _{\theta _\star }& = \partial _{\hat{\theta }_\star }+\partial _{\theta _\star }h\,\partial _\phi ,\\ \partial _{\phi _\star }& = \partial _\phi . \end{aligned}$$

These equations help us with the geometric interpretation of the change of coordinates operated by passing from \(\phi\) to \(\phi _\star\). Of course, for functions f that do not depend on \(\phi\), like the coefficients of our metric, \(\partial _uf=\partial _{\hat{u}}f\), \(\partial _vf=\partial _{\hat{v}}f\) and \(\partial _{\theta _\star }f=\partial _{\hat{\theta }_\star }f\). Defining

$$\begin{aligned} \Omega ^2& = -g_{r_{\star }r_{\star }}=-\,\frac{\rho ^2\Delta _r\Delta _\theta }{\Upsilon }, \end{aligned}$$
(19)
$$\begin{aligned} b^{\phi _{\star }}& = 2B=2\frac{\Xi a((r^2+a^2)\Delta _{\theta }-\Delta _r)}{\Upsilon }, \end{aligned}$$
(20)

the expression for the metric becomes

$$\begin{aligned} g& = -2\Omega ^2 \left( {\textrm{d}}u\otimes {\textrm{d}}v+{\,\textrm{d}}v\otimes {\textrm{d}}u\right) \nonumber \\{} & {} +{g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} {\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\theta _{\star } +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}\theta _{\star }\otimes ({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v) +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v)\otimes {\textrm{d}}\theta _{\star }\nonumber \\{} & {} +{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }} ({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v)\otimes ({\textrm{d}}\phi _{\star }-b^{\phi _{\star }}{\,\textrm{d}}v). \end{aligned}$$
(21)

For each pair (uv), \({g \hspace{-4.83322pt}/}\) is a metric defined on a two-sphere. The calculation above shows that the coefficients of this metric are

$$\begin{aligned} {g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }}& = g_{\theta _{\star }\theta _{\star }} +g_{\phi \phi }(\partial _{\theta _{\star }}h)^2 \ =\ \frac{\rho ^2L^2}{\Upsilon }+\frac{\sin ^2\theta }{\Xi ^2\rho ^2}\Upsilon (\partial _{\theta _{\star }}h)^2,\end{aligned}$$
(22)
$$\begin{aligned} {g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}& = g_{\phi \phi }(\partial _{\theta _{\star }}h) \ =\ \frac{\sin ^2\theta }{\Xi ^2\rho ^2}\Upsilon (\partial _{\theta _{\star }}h), \end{aligned}$$
(23)
$$\begin{aligned} {g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}& = g_{\phi \phi }\ =\ \frac{\sin ^2\theta }{\Xi ^2\rho ^2}\Upsilon . \end{aligned}$$
(24)

The determinants of the metrics \({g \hspace{-4.83322pt}/}\) and g are

$$\begin{aligned} \textrm{det}\,{g \hspace{-4.83322pt}/}=\frac{L^2\sin ^2\theta }{\Xi ^2} \end{aligned}$$
(25)

and

$$\begin{aligned} \hbox {det}\,g=- 4\Omega ^4 \frac{L^2 \sin ^2\theta }{\Xi ^2}, \end{aligned}$$

and the inverse of the metric g is

$$\begin{aligned} g^{-1}& = -\,\frac{1}{2\Omega ^2}\left( \partial _u\otimes (\partial _v+b^{\phi _{\star }}\, \partial _{\phi _{\star }})+(\partial _v+b^{\phi _{\star }}\, \partial _{\phi _{\star }})\otimes \partial _u\right) \\{} & {} +{g \hspace{-4.83322pt}/}^{\theta _{\star }\theta _{\star }}\,\partial _{\theta _{\star }}\otimes \partial _{\theta _{\star }} +{g \hspace{-4.83322pt}/}^{\theta _{\star }\phi _{\star }}\,\partial _{\theta _{\star }}\otimes \partial _{\phi _{\star }} +{g \hspace{-4.83322pt}/}^{\theta _{\star }\phi _{\star }}\,\partial _{\phi _{\star }}\otimes \partial _{\theta _{\star }} +{g \hspace{-4.83322pt}/}^{\phi _{\star }\phi _{\star }}\,\partial _{\phi _{\star }}\otimes \partial _{\phi _{\star }}, \end{aligned}$$

with coefficients given by

$$\begin{aligned} \left[ \begin{array}{ll} {g \hspace{-4.83322pt}/}^{\theta _{\star }\theta _{\star }}&{}{g \hspace{-4.83322pt}/}^{\theta _{\star }\phi _{\star }}\\ {g \hspace{-4.83322pt}/}^{\theta _{\star }\phi _{\star }}&{}{g \hspace{-4.83322pt}/}^{\phi _{\star }\phi _{\star }}\\ \end{array} \right] = \left[ \begin{array}{ll} \frac{\Upsilon }{L^2 \rho ^2} &{} -\,\frac{\Upsilon }{L^2 \rho ^2} (\partial _{\theta _{\star }}h) \\ -\,\frac{\Upsilon }{L^2 \rho ^2} (\partial _{\theta _{\star }}h)&{} \frac{\Xi ^2 \rho ^2}{\Upsilon \sin ^2\theta } +\frac{\Upsilon }{L^2 \rho ^2}(\partial _{\theta _{\star }}h)^2\\ \end{array} \right] . \end{aligned}$$

2.1.4 Normals to hypersurfaces and volume elements

We finish this subsection by writing down the volume elements of hypersurfaces

$$\begin{aligned} \Sigma _{r_\star }=\{r_\star =\hbox {constant}\},\quad \underline{\mathcal{C}}_v=\{v=\hbox {constant}\}\quad \hbox {and}\quad \mathcal{C}_u=\{u=\hbox {constant}\} \end{aligned}$$

of our spacetime, corresponding to constant \(r_\star\), v and u. As

$$\begin{aligned} 2\Omega ^2({\textrm{d}}u\otimes {\textrm{d}}v+{\,\textrm{d}}v\otimes {\textrm{d}}u)=\Omega ^2({\textrm{d}}r_\star \otimes {\textrm{d}}r_\star - {\textrm{d}}t\otimes {\textrm{d}}t), \end{aligned}$$

recalling (25) for the determinant of \({g \hspace{-4.83322pt}/}\), the volume element for \(\Sigma _{r_\star }\) is

$$\begin{aligned} \textrm{dVol}_{\Sigma _{r_\star }}=\Omega \frac{L \sin \theta }{\Xi }{\,\textrm{d}}\theta _{\star }{\,\textrm{d}}\phi _{\star } {\,\textrm{d}}t_{\star }. \end{aligned}$$

Our choice for the normals to constant v and u hypersurfaces are

$$\begin{aligned} n_{\underline{\mathcal{C}}_v}=\partial _u\qquad \hbox {and}\qquad n_{\mathcal{C}_u}=\partial _v+b^{\phi _{\star }}\partial _{\phi _{\star }}. \end{aligned}$$
(26)

We have that

$$\begin{aligned} g\left( \partial _u,\frac{1}{2\Omega ^2}\partial _v\right) =-1 \qquad \hbox {and}\qquad g\left( \partial _v+b^{\phi _{\star }}\partial _{\phi _{\star }},\frac{1}{2\Omega ^2}\partial _u\right) =-1 \end{aligned}$$

and so, since the volume element associated to the metric g is

$$\begin{aligned} \textrm{dVol}=2\Omega ^2 \frac{L \sin \theta }{\Xi } {\,\textrm{d}}u {\,\textrm{d}}v{\,\textrm{d}}\theta _{\star }{\,\textrm{d}}\phi _{\star }, \end{aligned}$$
(27)

the volume elements associated to constant v and u hypersurfaces are

$$\begin{aligned} \textrm{dVol}_{\underline{\mathcal{C}}_v}=\frac{L \sin \theta }{\Xi }{\,\textrm{d}}u{\,\textrm{d}}\theta _{\star }{\,\textrm{d}}\phi _{\star } \end{aligned}$$

and

$$\begin{aligned} \textrm{dVol}_{{\mathcal{C}_u}}=\frac{L \sin \theta }{\Xi }{\,\textrm{d}}v{\,\textrm{d}}\theta _{\star } {\,\textrm{d}}\phi _{\star }. \end{aligned}$$

As \(r_\star =u+v\), the tangent space to a hypersurface \(\Sigma _{r_\star }\), where \(r_\star\) is constant, is spanned by \(\partial _v-\partial _u\), \(\partial _{\theta _{\star }}\) and \(\partial _{\phi _{\star }}\), and

$$\begin{aligned} n_{\Sigma _{r_\star }}=\frac{\partial _u+\partial _v+b^{\phi _{\star }}\partial _{\phi _{\star }}}{2\Omega }=-\,\frac{\hbox {grad}\,r_\star }{(-g(\hbox {grad}\,r_\star ,\hbox {grad}\,r_\star ))^{\frac{1}{2}}}. \end{aligned}$$

Remark 2.3

\(\partial _{\phi _\star }\) is equal to zero when \(\theta _\star\) is either 0 or \(\pi\).

Indeed, the vector field \(\partial _{\phi _\star }\) is tangent to the spheres \(u=\hbox {constant}\) and \(v=\hbox {constant}\) which are contained in the spacelike hypersurfaces \(r_\star =\hbox {constant}\) and

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}g(\partial _{\phi _\star },\partial _{\phi _\star }) =\lim _{\theta _\star \rightarrow 0}{g \hspace{-4.83322pt}/}_{\phi _\star \phi _\star }=0. \end{aligned}$$

2.2 Regularity of the change of coordinates

The regularity of transformation of coordinates \((r_\star ,\theta _\star )\mapsto (r,\theta )\) for Kerr spacetimes, namely at \(\theta _\star =0\) and \(\theta _\star =\frac{\pi }{2}\), was shown by Dafermos and Luk [9]. In this subsection we adapt their work to the setting of Kerr–Newman–de Sitter spacetimes.

2.2.1 The coordinate \(\theta _{\star }\) is well defined and continuous

Lemma 2.4

\(\theta _{\star }\) is well defined.

Proof

Clearly

$$\begin{aligned} F(r,\theta ,\theta )\ge 0. \end{aligned}$$

Moreover

$$\begin{aligned} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}\ge & {} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{ a\sqrt{\Delta _{\theta '}-\sin ^2\theta '}}\nonumber \\& = \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{\Xi ^{\frac{1}{2}}a\cos \theta '}\rightarrow +\infty \ \ \hbox {as}\ \theta _{\star }\rightarrow \frac{\pi }{2}, \end{aligned}$$

which implies that

$$\begin{aligned} \lim _{\theta _{\star }\rightarrow \frac{\pi }{2}}F(r,\theta ,\theta _{\star })=-\infty . \end{aligned}$$

The continuity of F implies that for each pair \((r,\theta )\), with \(\theta \in \bigl (0,\frac{\pi }{2}\bigr )\), there exists \(\theta _{\star }=\theta _{\star }(r,\theta )\in \bigl (\theta ,\frac{\pi }{2}\bigr )\) such that \(F(r,\theta ,\theta _{\star })=0\).

The function F is decreasing in \(\theta _{\star }\). This follows from the fact that

$$\begin{aligned} \theta _{\star }\rightarrow \int _0^{\theta _\star }\frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}} \end{aligned}$$

is strictly increasing. Indeed, suppose \(\theta _{\star }<\tilde{\theta }_{\star }\). Setting \(\sin \tilde{\theta }'=\frac{\sin \tilde{\theta }_\star }{\sin \theta _{\star }}\sin \theta '\), as \(\theta '<\tilde{\theta }'\), we have \(\Delta _{\theta '}>\Delta _{\tilde{\theta }'}\) and

$$\begin{aligned}{} & {} \int _0^{\tilde{\theta }_\star }\frac{{\,\textrm{d}}\tilde{\theta }'}{\sqrt{\sin ^2 \tilde{\theta }_{\star }\Delta _{\tilde{\theta }'}-\sin ^2\tilde{\theta }'}} -\int _0^{\theta _\star }\frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}\\{} & {} \ \ \ge \int _0^{\tilde{\theta }_\star }\frac{{\,\textrm{d}}\tilde{\theta }'}{\sqrt{\sin ^2\tilde{\theta }_{\star }\Delta _{\theta '}-\sin ^2\tilde{\theta }'}} -\int _0^{\theta _\star }\frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}\\{} & {} \ \ =\int _0^{\theta _\star } \left( \frac{\cos \theta '}{\cos \tilde{\theta }'}-1\right) \frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}\ \ge \ 0. \end{aligned}$$

\(\square\)

Lemma 2.5

\(\theta _{\star }\) is continuous at \(\theta =0\) and \(\theta =\frac{\pi }{2}\) with

$$\begin{aligned} \frac{\sin \theta _{\star }}{\sin \theta }+\frac{\cos \theta }{\cos \theta _{\star }}\lesssim 1. \end{aligned}$$
(28)

Proof

The second integral in (9) is bounded above by

$$\begin{aligned} \int _{r}^{r_+}\frac{{\,\textrm{d}}r'}{Q(r',\theta _{\star })}\le & {} \int _{r_-}^{r_+} \frac{{\,\textrm{d}}r'}{r'^2+a^2}\nonumber \\& = \frac{1}{a}\left( \arctan \left( \frac{r_+}{a}\right) - \arctan \left( \frac{r_-}{a}\right) \right) \nonumber \\& = \frac{1}{a}\left( \frac{\pi }{2}-c_{\Lambda ,M,a,e}\right) , \end{aligned}$$
(29)

for some positive \(c_{\Lambda ,M,a,e}\). On the other hand, using the substitution \(\sin \theta '=\sin \theta _{\star }\sin \tilde{\theta }\) we see that the negative of the first integral in (9) is bounded below by

$$\begin{aligned} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}& = \frac{1}{a} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}- \sin ^2\theta '}}\nonumber \\& = \frac{1}{a}\int _{\arcsin \left( \frac{\sin \theta }{\sin \theta _{\star }}\right) }^{\frac{\pi }{2}} \frac{\cos \tilde{\theta }\,{\,\textrm{d}}\tilde{\theta }}{\cos \theta '\sqrt{\Delta _{\theta '}-\sin ^2\tilde{\theta }}}\nonumber \\& = \frac{1}{a}\int _{\arcsin \left( \frac{\sin \theta }{\sin \theta _{\star }}\right) }^{\frac{\pi }{2}} \frac{\cos \tilde{\theta }\,{\,\textrm{d}}\tilde{\theta }}{\sqrt{\left( 1+\frac{\Lambda }{3}a^2\right) -\left( 1+\frac{\Lambda }{3}a^2\sin ^2\theta _{\star }\right) \sin ^2\tilde{\theta }}}\nonumber \\\ge & {} \frac{1}{a}\int _{\frac{\sin \theta }{\sin \theta _{\star }}}^{1}\frac{{\,\textrm{d}}x}{\sqrt{\Xi -x^2}}\nonumber \\& = \frac{1}{a}\int _{\frac{\sin \theta }{\sqrt{\Xi }\sin \theta _{\star }}}^{ \frac{1}{\sqrt{\Xi }}}\frac{{\,\textrm{d}}y}{\sqrt{1-y^2}}\nonumber \\& = \frac{1}{a}\left( \arcsin \left( \frac{1}{\sqrt{\Xi }}\right) -\, \arcsin \left( \frac{\sin \theta }{\sqrt{\Xi }\sin \theta _{\star }}\right) \right) \nonumber \\\ge & {} \frac{1}{a}\left( \arcsin \left( \frac{1}{\sqrt{\Xi }}\right) -\, \arcsin \left( \frac{\sin \theta }{\sin \theta _{\star }}\right) \right) . \end{aligned}$$
(30)

Since \(F(r,\theta ,\theta _{\star })\) is equal zero, we obtain the estimate

$$\begin{aligned} \arcsin \left( \frac{\sin \theta }{\sin \theta _{\star }}\right) \ge \arcsin \left( \frac{1}{\sqrt{\Xi }}\right) +c_{\Lambda ,M,a,e}-\,\frac{\pi }{2}, \end{aligned}$$
(31)

provided the right-hand side is positive, i.e.

$$\begin{aligned} \Xi< & {} \csc ^2\left( \frac{\pi }{2}-c_{\Lambda ,M,a,e}\right) \nonumber \\& = \csc ^2 \left( \arctan \left( \frac{r_+}{a}\right) - \arctan \left( \frac{r_-}{a}\right) \right) . \end{aligned}$$
(32)

For fixed \(r_-\) and \(r_+\), the minimum of the last expression is attained when \(a=\sqrt{r_-r_+}\). So the last inequality is implied by the stronger restriction

$$\begin{aligned} \Xi <\csc ^2\left( \arctan \left( \sqrt{\alpha }\right) - \arctan \left( \frac{1}{\sqrt{\alpha }}\right) \right) ,\qquad \alpha =\frac{r_+}{r_-}. \end{aligned}$$
(33)

According to Lemma B.1, for all subextremal black holes, we have

$$\begin{aligned} \Xi <1+\frac{l(\alpha ,\Lambda e^2)}{3}\le 1+\frac{l(\alpha ,0)}{3}, \end{aligned}$$

where l is given by (138), so that

$$\begin{aligned} l(\alpha ,0):& = \frac{3}{2 \alpha ^2 (1+2 \alpha )} \left( \left( 1+4 \alpha +12 \alpha ^2+16 \alpha ^3+9 \alpha ^4\right) \right. \nonumber \\{} & {} \left. \qquad \qquad -\left( 1+3 \alpha +5 \alpha ^2+3 \alpha ^3\right) \sqrt{1+2\alpha +9\alpha ^2}\right) . \end{aligned}$$

We claim that

$$\begin{aligned} 1+\frac{l(\alpha ,0)}{3}<\csc ^2\left( \arctan \left( \sqrt{\alpha }\right) - \arctan \left( \frac{1}{\sqrt{\alpha }}\right) \right) =\left( \frac{\alpha +1}{\alpha -1}\right) ^2, \end{aligned}$$

for all \(\alpha\) greater than 1. To show this, we define the function

$$\begin{aligned} \hat{\rho }(\alpha ):=\frac{\bigl (\frac{\alpha +1}{\alpha -1}\bigr )^2-1}{\frac{l(\alpha ,0)}{3}}= \frac{\frac{4\alpha }{(\alpha -1)^2}}{\frac{l(\alpha ,0)}{3}}, \end{aligned}$$

which we wish to check is always greater than one. By direct computation one may check that

$$\begin{aligned} l(\alpha ,0)\le \frac{2}{3\alpha }. \end{aligned}$$

Indeed,

$$\begin{aligned} l(\alpha ,0)\le \frac{2}{3\alpha }\Leftrightarrow & {} \left( 1+4 \alpha +12 \alpha ^2+16 \alpha ^3+9 \alpha ^4\right) -\left( 1+3 \alpha +5 \alpha ^2+3 \alpha ^3\right) \sqrt{1+2\alpha +9\alpha ^2}\le \frac{4}{9}\alpha (1+2\alpha )\\\Leftrightarrow & {} 1+\frac{32\alpha }{9}+\frac{100\alpha ^2}{9}+16\alpha ^3+9\alpha ^4\le \left( 1+3 \alpha +5 \alpha ^2+3 \alpha ^3\right) \sqrt{1+2\alpha +9\alpha ^2}. \end{aligned}$$

Since both sides of the last inequality are positive, the inequality holds if and only if the square of the right-hand side minus the square of the left-hand side is nonnegative. Calculating this difference, one finds out that the terms in \(\alpha ^8\), \(\alpha ^7\) and \(\alpha ^6\) cancel out. One is left with the polynomial

$$\begin{aligned} \frac{4\alpha }{81}(18+104\alpha +344\alpha ^2+623\alpha ^3+414\alpha ^4), \end{aligned}$$

whose coefficients are all positive. Hence, this polynomial is positive (for \(\alpha >1\)). So,

$$\begin{aligned} \hat{\rho }(\alpha )\ge \frac{18\alpha ^2}{(\alpha -1)^2}\ge 18>1. \end{aligned}$$

The claim is proven.

The previous paragraph implies that (33) is satisfied, and so the right-hand side of (31) is positive. Then, we have

$$\begin{aligned} \frac{\sin \theta }{\sin \theta _{\star }}\ge c>0. \end{aligned}$$
(34)

If we use the substitution \(\cos \theta '=\cos \theta _{\star }\sec \tilde{\theta }\) in (30) we get

$$\begin{aligned} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}& = \frac{1}{a} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}- \sin ^2\theta '}}\\& = \frac{1}{a}\int _0^{{{\,\textrm{arcsec}\,}}\left( \frac{\cos \theta }{\cos \theta _{\star }}\right) } \frac{\sec \tilde{\theta }\cos \theta _\star \tan \tilde{\theta }{\,\textrm{d}}\tilde{\theta }}{\sin \theta '\sqrt{\Delta _{\theta '}-1+\cos ^2\theta _{\star }\sec ^2\tilde{\theta }-\Delta _{\theta '}\cos ^2\theta _{\star }}}\\& = \frac{1}{a}\int _0^{{{\,\textrm{arcsec}\,}}\left( \frac{\cos \theta }{\cos \theta _{\star }}\right) } \frac{\sec \tilde{\theta }\tan \tilde{\theta }{\,\textrm{d}}\tilde{\theta }}{\sin \theta '\sqrt{\frac{\Lambda }{3}a^2\sec ^2\tilde{\theta }+\sec ^2\tilde{\theta }-\Delta _{\theta '}}}\\& = \frac{1}{a}\int _0^{{{\,\textrm{arcsec}\,}}\left( \frac{\cos \theta }{\cos \theta _{\star }}\right) } \frac{\sec \tilde{\theta }\tan \tilde{\theta }{\,\textrm{d}}\tilde{\theta }}{\sin \theta '\sqrt{\frac{\Lambda }{3}a^2\sec ^2\tilde{\theta }\sin ^2\theta _{\star }+\tan ^2\tilde{\theta }}}\\& = \frac{1}{a}\int _0^{{{\,\textrm{arcsec}\,}}\left( \frac{\cos \theta }{\cos \theta _{\star }}\right) } \frac{\sec \tilde{\theta }{\,\textrm{d}}\tilde{\theta }}{\sin \theta '\sqrt{1+\frac{\Lambda }{3}a^2\frac{\sin ^2\theta _{\star }}{\sin ^2\tilde{\theta }}}}. \end{aligned}$$

If \(\frac{\cos \theta }{\cos \theta _{\star }}\le \sqrt{2}\), we are done. Otherwise, as

$$\begin{aligned} \frac{1}{\sin ^2({{\,\textrm{arcsec}\,}}x)}=\frac{1}{1-\,\frac{1}{x^2}} \end{aligned}$$

and \(\sin ^2\theta _{\star }\le 1\), we have that

$$\begin{aligned} \int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}\ge & {} \frac{1}{a}\int _{\frac{\pi }{4}}^{{{\,\textrm{arcsec}\,}}\left( \frac{\cos \theta }{\cos \theta _{\star }}\right) } \frac{\sec \tilde{\theta }{\,\textrm{d}}\tilde{\theta }}{\sqrt{1+2\frac{\Lambda }{3}a^2}}\\\ge & {} \frac{1}{\sqrt{2}\Xi ^\frac{1}{2} a} \ln (\sec x+\tan x)\Big |_{\frac{\pi }{4}}^{{{\,\textrm{arcsec}\,}}\left( \frac{\cos \theta }{\cos \theta _{\star }}\right) }\\& = \frac{1}{\sqrt{2}\Xi ^\frac{1}{2} a}\left[ \ln \left( \frac{\cos \theta }{\cos \theta _{\star }}+ \sqrt{\frac{\cos ^2\theta }{\cos ^2\theta _{\star }}-1}\right) -\ln (1+\sqrt{2}) \right] . \end{aligned}$$

The result follows since (29) implies that the last expression is bounded above. \(\square\)

2.2.2 The derivative of the function defining \(\theta _\star\)

Remark 2.6

Let us define

$$\begin{aligned} D(\theta ,\theta _{\star }):=\sin ^2\theta _{\star } \Delta _\theta -\sin ^2\theta , \end{aligned}$$

so that \(D(\theta ,\theta _{\star })=P^2(\theta ,\theta _{\star })/a^2\). Clearly,

$$\begin{aligned} \partial _\theta D(\theta ,\theta _{\star })= \partial _\theta \left( \sin ^2\theta _{\star } \left( 1+\frac{\Lambda }{3} a^2 \cos ^2\theta \right) -\sin ^2\theta \right) =-\sin (2\theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \end{aligned}$$

holds. The trigonometric equality

$$\begin{aligned} \sin ^2(2 \theta _{\star }) \left( 1+\frac{\Lambda }{3} a^2 \cos ^2\theta \right)& = \sin ^2(2 \theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \\{} & {} +2 (\cos (2 \theta )+\cos (2 \theta _{\star })) \left( \sin ^2\theta _{\star } \left( 1+\frac{\Lambda }{3} a^2 \cos ^2\theta \right) -\sin ^2\theta \right) \end{aligned}$$

allows us to conclude that

$$\begin{aligned} \sin ^2(2 \theta _{\star }) \Delta _\theta& = -\sin (2 \theta ) \left[ \partial _\theta \left( \sin ^2\theta _{\star } \Delta _\theta -\sin ^2\theta \right) \right] \nonumber \\{} & {} +2 (\cos (2 \theta )+\cos (2 \theta _{\star })) \left( \sin ^2\theta _{\star } \Delta _\theta -\sin ^2\theta \right) \nonumber \\& = -\sin (2\theta )\,\partial _\theta D(\theta ,\theta _{\star })+ 2 (\cos (2 \theta )+\cos (2 \theta _{\star }))D(\theta ,\theta _{\star }). \end{aligned}$$
(35)

We will have to use the identity (35) repeatedly in our calculations.

Lemma 2.7

For \(\theta _{\star }\in \left( 0,\frac{\pi }{2}\right)\), we have that

$$\begin{aligned} G(r,\theta ,\theta _{\star })& = \partial _{\theta _{\star }} F(r,\theta ,\theta _{\star }) \nonumber \\& = -\,\frac{\csc (2\theta _{\star })\sin (2\theta )}{a\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }} -\,\frac{2\csc (2\theta _{\star })}{a} \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}{\,\textrm{d}}\theta '\nonumber \\{} & {} +\int _r^{r_+} \frac{a^2\sin (2\theta _{\star }) \Delta _{r'}}{2\left( (r'^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _{r'}\right) ^{\frac{3}{2}}}{\,\textrm{d}}r'. \end{aligned}$$
(36)

Proof

We start from the definition of F in (9). Since \(\theta _{\star }\in \left( 0,\frac{\pi }{2}\right)\), for \(\theta '\in [\theta ,\theta _{\star }]\), we get \(\Delta _{\theta '}\ge 1+\frac{\Lambda }{3}a^2\cos ^2\theta _{\star }\), and so \(P(\theta ',\theta _{\star })\ge \sqrt{\frac{\Lambda }{3}}\frac{a^2}{2}\sin {(2\theta _{\star })}>0\). So, we do not have to worry about vanishing denominators. The result follows from the calculation

$$\begin{aligned}{} & {} a\frac{\partial }{\partial \theta _{\star }}\Bigg |_{\theta \tiny \hbox {fixed}}\int _{\theta }^{\theta _{\star }}\frac{{\,\textrm{d}}\theta '}{P(\theta ', \theta _{\star })}=\frac{\partial }{\partial \theta _{\star }}\Bigg |_{\theta \tiny \hbox {fixed}}\int _{\theta }^{\theta _{\star }} \frac{{\,\textrm{d}}\theta '}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}\\{} & {} \ \ = \frac{2}{\sqrt{\frac{\Lambda }{3}}a\sin (2\theta _{\star })}- \int _{\theta }^{\theta _{\star }} \frac{\sin (2\theta _{\star })\Delta _{\theta '}}{2\sqrt{\left( D(\theta ',\theta _{\star })\right) ^3}}{\,\textrm{d}}\theta '\\{} & {} \ \ {\mathop {=}\limits ^{(35)}} \frac{2}{\sqrt{\frac{\Lambda }{3}}a\sin (2\theta _{\star })}+ \int _{\theta }^{\theta _{\star }} \frac{\csc (2\theta _{\star })\sin (2\theta ') \partial _{\theta '}D(\theta ',\theta _{\star })}{2\sqrt{\left( D(\theta ',\theta _{\star })\right) ^3}}{\,\textrm{d}}\theta '\\{} & {} \ \ \ \ \ \ - \int _{\theta }^{\theta _{\star }} \frac{\csc (2\theta _{\star })(\cos (2\theta ')+\cos (2\theta _{\star }))}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '\\{} & {} \ \ = \frac{2}{\sqrt{\frac{\Lambda }{3}}a\sin (2\theta _{\star })}+ \int _{\theta }^{\theta _{\star }} \frac{\csc (2\theta _{\star })\sin (2\theta ') \partial _{\theta '}D(\theta ',\theta _{\star })}{2\sqrt{\left( D(\theta ',\theta _{\star })\right) ^3}}{\,\textrm{d}}\theta '\\{} & {} \ \ \ \ \ \ - \int _{\theta }^{\theta _{\star }} \frac{2\csc (2\theta _{\star })\cos (2\theta ')}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '+ \int _{\theta }^{\theta _{\star }} \frac{\csc (2\theta _{\star })(\cos (2\theta ')-\cos (2\theta _{\star }))}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '\\{} & {} \ \ = \frac{2}{\sqrt{\frac{\Lambda }{3}}a\sin (2\theta _{\star })}- \int _{\theta }^{\theta _{\star }}\frac{\partial }{\partial \theta '} \frac{\csc (2\theta _{\star })\sin (2\theta ')}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '\\{} & {} \ \ \ \ \ \ + \int _{\theta }^{\theta _{\star }} \frac{2\csc (2\theta _{\star })(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '\\{} & {} \ \ = \frac{\csc (2\theta _{\star })\sin (2\theta )}{\sqrt{D(\theta ,\theta _{\star })}} +2\csc (2\theta _{\star }) \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '. \end{aligned}$$

For the second equality we used (35). \(\square\)

The expression for G and the definition of L in (14) yield

$$\begin{aligned} L& = \sqrt{(r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r}\Biggl (\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\Biggr .\nonumber \\{} & {} +2\sin ^2(2\theta _{\star })\frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta } -\sin ^2\theta }}{\sin (2\theta _{\star })}\left( \frac{1}{\sin ^2(2\theta _{\star })} \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}{\,\textrm{d}}\theta '\right) \nonumber \\{} & {} \Biggl . -\sin ^2(2\theta _{\star })\frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2\theta _{\star })}\int _r^{r_+} \frac{a^3 \Delta _{r'}}{2\left( (r'^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _{r'}\right) ^{\frac{3}{2}}}{\,\textrm{d}}r'\Biggr ). \end{aligned}$$
(37)

Lemma 2.8

We have the following estimate for G:

$$\begin{aligned} -\,\frac{C_{\Lambda ,M,a,e}}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}\le G(r,\theta , \theta _{\star }(r,\theta ))\le -\,\frac{\csc (2\theta _{\star })\sin (2\theta )}{ a\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}. \end{aligned}$$
(38)

for some constant \(C_{\Lambda ,M,a,e}>0\) depending on \(\Lambda , M, a\) and e.

Proof

We use expression (36). Note that

$$\begin{aligned} \frac{2\csc (2\theta _{\star })}{a} \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}{\,\textrm{d}}\theta ' \le \frac{2\csc (2\theta _{\star })}{a} \int _{\theta }^{\theta _{\star }} \sqrt{\sin ^2\theta _{\star }-\sin ^2\theta '}{\,\textrm{d}}\theta '. \end{aligned}$$

For \(\theta _{\star }\) close to zero, the right-hand side is bounded above by

$$\begin{aligned} \frac{\pi /2}{a\cos \theta _{\star }}, \end{aligned}$$

whereas for \(\theta _{\star }\) close to \(\frac{\pi }{2}\), the right-hand side is bounded above by

$$\begin{aligned} \frac{2\csc (2\theta _{\star })}{a}\int _{\theta }^{\theta _{\star }}\cos \theta '{\,\textrm{d}}\theta '& = \frac{2\csc (2\theta _{\star })}{a}(\sin \theta _{\star }-\sin \theta )\\\le & {} \frac{2\csc (2\theta _{\star })}{a}(\theta _{\star }-\theta )\cos \theta \\\le & {} \frac{\pi /2}{a\sin \theta _{\star }}\frac{\cos \theta }{\cos \theta _\star }. \end{aligned}$$

Recalling (28), we conclude that the second term on the right-hand side of (36) is uniformly bounded below by a negative constant. Clearly, the same is true for the third term on the right-hand side of (36). Since both terms are negative, \(\csc (2\theta _{\star })\sin (2\theta )\) is bounded, and \(1/\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }\) is bounded below by a positive constant, we obtain (38). \(\square\)

2.2.3 The coordinates r and \(\theta\) as functions of \(r_\star\) and \(\theta _\star\)

Lemma 2.9

The mapping \((r,\theta )\rightarrow (r_\star ,\theta _\star )\), from \((r_-,r_+)\times \left[ 0,\frac{\pi }{2}\right]\) to \((-\infty ,+\infty )\times \left[ 0,\frac{\pi }{2}\right]\), is globally invertible.

Proof

Fix \((r_\star )_0\in {{\mathbb {R}}}\). For each fixed \(\theta \in \bigl (0,\frac{\pi }{2}\bigr )\), let \(r(\theta )\) be the unique (because \(\partial _rr_\star <0\)) value of \(r\in (r_-,r_+)\) such that

$$\begin{aligned} r_\star (r(\theta ),\theta )=(r_\star )_0. \end{aligned}$$

The Implicit Function Theorem guarantees that \((r(\,\cdot \,),\,\cdot \,)\) is a \(C^1\) curve and that

$$\begin{aligned} \dot{r}=-\,\frac{\partial _\theta r_\star }{\partial _rr_\star }=-\,\frac{P}{Q} \frac{\Delta _r}{\Delta _\theta }. \end{aligned}$$

To see how \(\theta _\star\) varies along this curve we calculate

$$\begin{aligned} \frac{d}{d\theta }\theta _\star (r(\theta ),\theta )=\dot{r}\,\partial _r\theta _\star + \partial _\theta \theta _\star = \frac{1}{GQ}\left( -\,\frac{P}{Q} \frac{\Delta _r}{\Delta _\theta }\right) -\,\frac{1}{GP}=-\,\frac{\Upsilon }{GPQ^2}>0. \end{aligned}$$

So, \(\theta _\star\) is strictly increasing along the curve \((r(\,\cdot \,),\,\cdot \,)\). Moreover, (28) shows that \(\lim _{\theta \searrow 0}\theta _\star (r(\theta ),\theta )=0\) and \(\lim _{\theta \nearrow \frac{\pi }{2}}\theta _\star (r(\theta ),\theta )=\frac{\pi }{2}\). Thus, given \(((r_\star )_0,(\theta _\star )_0)\in (-\infty ,+\infty )\times \left[ 0,\frac{\pi }{2}\right]\) there exists one and only one \((r,\theta )\) such that \((r_\star (r,\theta ),\theta _\star (r,\theta ))= ((r_\star )_0,(\theta _\star )_0)\). We conclude that the mapping

$$\begin{aligned} (r_\star ,\theta _\star ):(r_-,r_+)\times \left[ 0,\frac{\pi }{2}\right] \longrightarrow (-\infty ,+\infty )\times \left[ 0,\frac{\pi }{2}\right] \end{aligned}$$

is one-to-one and onto. \(\square\)

2.2.4 First partial derivatives

Lemma 2.10

The partial derivatives of \((r, {\theta })\) with respect to \((r_{\star }, \theta _{\star })\) are given by

$$\begin{aligned} \frac{\partial r}{\partial r_{\star }}& = \frac{\Delta _r \Delta _{\theta }Q}{\Upsilon }=\frac{\Delta _r \Delta _{\theta } \sqrt{(r^2+a^2)^2-a^2 \Delta _r \sin ^2{\theta _{\star }}}}{\left( (r^2+a^2)^2 \Delta _{\theta }-a^2 \Delta _r \sin ^2{\theta }\right) }, \end{aligned}$$
(39)
$$\begin{aligned} \frac{\partial r}{\partial {\theta _{\star }}}& = -\,\frac{\Delta _rL P }{\Upsilon }= \frac{\Delta _rG P^2 Q}{\Upsilon }=G \frac{a^2 \Delta _r \left( \sin ^2{\theta _{\star }}\Delta _{\theta } -\sin ^2{\theta }\right) \sqrt{(r^2+a^2)^2-a^2 \sin ^2{\theta _{\star }}\Delta _r}}{(r^2+a^2)^2 \Delta _{\theta }-a^2\sin ^2{\theta } \Delta _r },\nonumber \\& = \frac{a\Delta _r \sqrt{(r^2+a^2)^2-a^2 \sin ^2{\theta _{\star }}\Delta _r}}{(r^2+a^2)^2 \Delta _{\theta }-a^2\sin ^2{\theta } \Delta _r } \Biggl ( -\,\frac{\sin (2\theta )}{\sin (2\theta _{\star })} \frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2\theta _{\star })}\Biggr .\nonumber \\{} & {} -2\frac{\left( \sin ^2{\theta _{\star }}\Delta _{\theta } -\sin ^2{\theta }\right) }{\sin ^2(2\theta _{\star })} \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}{\,\textrm{d}}\theta ' \nonumber \\{} & {} \Biggl .+\frac{\left( \sin ^2{\theta _{\star }}\Delta _{\theta } -\sin ^2{\theta }\right) }{\sin ^2(2\theta _{\star })} \int _r^{r_+} \frac{a^3\sin ^2(2\theta _{\star }) \Delta _{r'}}{2\left( (r'^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _{r'}\right) ^{\frac{3}{2}}}{\,\textrm{d}}r'\Biggr )\sin (2\theta _{\star }), \end{aligned}$$
(40)
$$\begin{aligned} \frac{\partial {\theta }}{\partial r_{\star }}& = \frac{\Delta _r \Delta _{\theta }P}{\Upsilon }= \frac{\sqrt{ \sin ^2{\theta _{\star }}\Delta _{\theta }-\sin ^2{\theta }}}{\sin (2\theta _{\star })} \frac{a \Delta _r \Delta _{\theta }}{\left( (r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2{\theta }\Delta _r\right) }\sin (2\theta _{\star }), \end{aligned}$$
(41)
$$\begin{aligned} \frac{\partial {\theta }}{\partial {\theta _{\star }}}& = \frac{\Delta _{\theta }L Q }{\Upsilon } =-\,\frac{\Delta _{\theta }G P Q^2 }{\Upsilon }=-G\frac{a \Delta _{\theta } \sqrt{ \sin ^2{\theta _{\star }}\Delta _{\theta }-\sin ^2{\theta }} \left( (r^2+a^2)^2-a^2 \sin ^2{\theta _{\star }}\Delta _r \right) }{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2{\theta }\Delta _r}\nonumber \\& = \Delta _{\theta }\frac{(r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r}{ (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}}\Biggl ({\frac{\sin (2\theta )}{\sin (2\theta _{\star })}} \Biggr .\nonumber \\{} & {} \ \ \ \ +\,{2\frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2\theta _{\star })}} \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}{\,\textrm{d}}\theta ' \nonumber \\{} & {} \Biggl .\ \ \ \ -\,\frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2\theta _{\star })}\int _r^{r_+} \frac{a^3\sin ^2(2\theta _{\star }) \Delta _{r'}}{2\left( (r'^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _{r'}\right) ^{\frac{3}{2}}}{\,\textrm{d}}r'\Biggr ). \end{aligned}$$
(42)

In the region \(r_-<r<r_+\), we have

$$\begin{aligned} \begin{array}{ll} \displaystyle \left| \frac{\partial r}{\partial r_{\star }}\right| \lesssim |\Delta _r|, &{}\displaystyle \left| \frac{\partial r}{\partial {\theta _{\star }}}\right| \lesssim |\Delta _r|\sin (2 {\theta _{\star }})\\ \\ \displaystyle \left| \frac{\partial {\theta }}{\partial r_{\star }}\right| \lesssim |\Delta _r|\sin (2 {\theta _{\star }}), &{}\displaystyle \left| \frac{\partial {\theta }}{\partial {\theta _{\star }}}\right| \lesssim 1. \end{array} \end{aligned}$$
(43)

Proof

We start from (16) and (17). We use (36) to obtain (40) and (42). The estimates for the derivatives of r and \(\theta\) follow from the estimates

$$\begin{aligned} \sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }\le \sin \theta _{\star }\sqrt{\Delta _{\theta }}\lesssim \sin \theta _{\star } \lesssim \sin (2\theta _{\star }),\ \hbox {for}\ \theta _{\star }\ \hbox {close to}\ 0,\\ \sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }= \cos \theta _{\star }\sqrt{\Xi \frac{\cos ^2\theta }{\cos ^2\theta _{\star }}-\Delta _\theta } {\mathop {\lesssim }\limits ^{(28)}}\cos \theta _{\star }\lesssim \sin (2\theta _{\star }), \ \hbox {for}\ \theta _{\star }\ \hbox {close to}\ \frac{\pi }{2}, \end{aligned}$$

which together imply that

$$\begin{aligned} \frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2\theta _{\star })} \lesssim 1. \end{aligned}$$

\(\square\)

2.2.5 Higher order derivatives

Lemma 2.11

The functions r and \(\frac{\partial \theta }{\partial \theta _{\star }}\) are \(C^\infty\). For every \(k\ge 2\), we have that

$$\begin{aligned} \sum _{1\le k_1+k_2\le k}\left| \left( \partial _{r_{\star }}\right) ^{k_1}\left( \frac{1}{\sin (2\theta _{\star })} \partial _{\theta _{\star }}\right) ^{k_2}r\right|\lesssim & {} |\Delta _r|, \end{aligned}$$
(44)
$$\begin{aligned} \sum _{1\le k_1\le k}\left| \left( \partial _{r_{\star }}\right) ^{k_1}\theta \right|\lesssim & {} |\Delta _r| \sin (2\theta _{\star }), \end{aligned}$$
(45)
$$\begin{aligned} \sum _{0\le k_1\le k-1}\left| \left( \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\right) ^{k_1}\left( \frac{\partial \theta }{\partial \theta _{\star }}\right) \right|\lesssim & {} 1, \end{aligned}$$
(46)
$$\begin{aligned} \sum _{{\mathop { {\tiny { k_1 \ge 1}}}\limits ^{{1\le k_1+k_2\le k-1}}}} \left| \left( \partial _{r_{\star }}\right) ^{k_1} \left( \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\right) ^{k_2} \left( \frac{\partial \theta }{\partial \theta _{\star }}\right) \right|\lesssim & {} |\Delta _r|. \end{aligned}$$
(47)

Moreover, the derivatives of the function L given in (14) are bounded as follows:

$$\begin{aligned} \sum _{0\le k_1\le k-1}\left| \left( \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\right) ^{k_1}L\right|\lesssim & {} 1, \end{aligned}$$
(48)
$$\begin{aligned} \sum _{{\mathop { {\tiny { k_1 \ge 1}}}\limits ^{{1\le k_1+k_2\le k-1}}}} \left| \left( \partial _{r_{\star }}\right) ^{k_1} \left( \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\right) ^{k_2} L\right|\lesssim & {} |\Delta _r|. \end{aligned}$$
(49)

Remark 2.12

We say that a function depending on \(r_\star\) and \(\theta _\star\) is \(C^\infty\) if it has derivatives of all orders with respect of \(\partial _{r_\star }\) and \(\frac{1}{\sin (2\theta _\star )}\partial _{\theta _\star }\). Refer to Remarks (2.19), (2.20) and (2.21) for an explanation about the reason for introducing the factor \(\frac{1}{\sin (2\theta _\star )}\) behind \(\partial _{\theta _\star }\). The function \(\sin ^2\theta\) is \(C^\infty\).

To prove Lemma 2.11 we need to know the derivatives of

$$\begin{aligned} R_1& = \frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '},\\ R_2& = \frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})},\\ R_3& = \frac{\sin ^2\theta _{\star }-\sin ^2\theta }{\sin ^2 (2 {\theta _{\star }})},\\ R_4& = \frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}. \end{aligned}$$

Note that the dependence of \(\frac{\partial r}{\partial r_{\star }}\), \(\frac{1}{\sin (2\theta _{\star })}\frac{\partial r}{\partial \theta _{\star }}\), \(\frac{\partial \theta }{\partial \theta _{\star }}\) and L on \(\theta\) and \(\theta _{\star }\) is done through \(\sin ^2\theta\), \(\sin ^2\theta _{\star }\), \(\Delta _\theta\) (which is a function of \(\sin ^2\theta\), and this is crucial for our argument to work), \(\sin ^2(2\theta _{\star })\), and \(R_1\) to \(R_4\). The same is true for \(\frac{\partial \theta }{\partial r_{\star }}\), which however also has a factor \(\sin (2\theta _{\star })\). Moreover, note that

$$\begin{aligned} \begin{array}{rcl} \partial _{r_{\star }}(\sin ^2\theta )&{}=&{}\displaystyle \sin (2\theta )\frac{\partial \theta }{\partial r_{\star }},\quad \hbox {this has a factor}\ \sin (2\theta )\sin (2\theta _{\star }),\\ \displaystyle \partial _{r_{\star }}(\sin (2\theta )\sin (2\theta _{\star }))&{}=&{}\displaystyle \cos (2\theta )\frac{\partial \theta }{\partial r_{\star }}\sin (2\theta _{\star }),\quad \hbox {this has factors}\ \cos (2\theta )\ \hbox {and}\ \sin ^2(2\theta _{\star }),\\ \displaystyle \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}(\sin (2\theta )\sin (2\theta _{\star }))&{}=&{}\displaystyle 2\cos (2\theta ) \frac{\partial \theta }{\partial \theta _{\star }}+2\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\cos (2\theta _{\star })\\ \displaystyle \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}(\sin ^2\theta )&{}=&{}\displaystyle \frac{\sin (2\theta )}{\sin (2\theta _{\star })}\frac{\partial \theta }{\partial \theta _{\star }},\\ \displaystyle \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}(\sin ^2\theta _{\star }) &{}=&{}1,\\ \displaystyle \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}(\sin ^2(2\theta _{\star }))&{}=&{} 4\cos (2\theta _{\star }),\\ \displaystyle \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\cos (2\theta )&{}=&{}\displaystyle -2\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\frac{\partial \theta }{\partial \theta _{\star }},\\ \displaystyle \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\cos (2\theta _{\star }) &{}=&{}-2, \end{array} \end{aligned}$$
(50)

or, alternatively, \(\cos (2\theta )=1-2\sin ^2\theta\). If the factor \(\sin (2\theta )\) were to appear by itself in our formulas, or if the factor \(\sin (2\theta _{\star })\) were to appear by itself, then our argument would not go through because the derivatives \(\frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\) of each of these are not bounded. The structure of our problem is such that when one of these factors is present, then the other one is also present, and their product has derivatives of all orders with respect to \(\partial _{r_{\star }}\) and \(\frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\) which are bounded. Furthermore, as we will see below, the derivatives of \(R_1\) to \(R_4\) are sums whose summands are products of factors that are either \(R_1\), \(R_2\), \(R_3\), \(R_4\), or, when this is not the case, others that are clearly smooth (since the denominators that will appear do not vanish for \(r_-\le r\le r_+\)). Hence, it turns out that to prove that r and \(\theta\) are smooth, we just have to check that \(R_1\) to \(R_4\) are \(C^1\) and that their first derivatives have the aforementioned property. Next we calculate the first derivatives of \(R_1\) to \(R_4\).

As a small note, let \(\theta _{\star }\in \bigl (0,\frac{\pi }{2}\bigr )\). Consider the function

$$\begin{aligned} \frac{1}{\sin (2 {\theta _{\star }})}\int _\theta ^{\theta _{\star }}{\,\textrm{d}}\theta ' = \frac{{\theta _{\star }}-{\theta (r_{\star },\theta _{\star })}}{\sin (2 {\theta _{\star }})}. \end{aligned}$$

This is clearly bounded for \(\theta _{\star }\) close to zero. It is also bouded for \({\theta _{\star }}\) close to \(\frac{\pi }{2}\). Indeed, in this situation, we have

$$\begin{aligned} \frac{{\theta _{\star }}-{\theta }}{\sin (2 {\theta _{\star }})}\le \frac{\frac{\pi }{2}-{\theta }}{\sin (2 {\theta _{\star }})}=\frac{\frac{\pi }{2}-{\theta }}{2 \sin ({\theta _{\star }}) \cos ({\theta _{\star }})}\le \frac{C \left( \frac{\pi }{2}-{\theta }\right) }{\cos ({\theta })}=\frac{C \left( \frac{\pi }{2}-{\theta }\right) }{\sin \left( \frac{\pi }{2}-{\theta }\right) }\le C, \end{aligned}$$

because \(\frac{\cos \theta }{\cos \theta _{\star }}\) is bounded ((28)). So the function is bounded for \(\theta _{\star }\in \bigl (0,\frac{\pi }{2}\bigr )\).

Remark 2.13

The reader will notice, using the expressions below, that

$$\begin{aligned} |\partial _{r_{\star }}R_i|\lesssim & {} |\Delta _r|,\end{aligned}$$
(51)
$$\begin{aligned} \left| \frac{1}{\sin (2 {\theta _{\star }})} \partial _{\theta _{\star }}R_i\right|\lesssim & {} 1, \end{aligned}$$
(52)

for each \(i\in \{1,2,3,4\}\).

Derivatives of \(R_1\). For any integer \(n\ge 1\), the following identities hold:

$$\begin{aligned}{} & {} \partial _{r_{\star }}\left( \frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}\right) \nonumber \\{} & {} \qquad =-\,\frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n}{\sin ^{2 n}(2 {\theta _{\star }})}\frac{a \Delta _r \Delta _{\theta }}{\left( (r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r\right) },\end{aligned}$$
(53)
$$\begin{aligned}{} & {} \left( \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\right) \left( \frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}\right) \nonumber \\{} & {} \ \ \ \ =\frac{2 (2 n+1)}{\sin ^{2 (n+1)}(2 {\theta _{\star }})} \int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^{n+1}}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}\nonumber \\{} & {} \ \ \ \qquad -\Biggl (\frac{2}{\sin ^2 (2 {\theta _{\star }})} \int _{\theta }^{\theta _{\star }} \frac{\sin ^2\theta _{\star }-\sin ^2\theta '}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}\Biggr .\nonumber \\{} & {} \qquad \ \ \qquad \Biggl . - \int _r^{r_+} \frac{a^3 \Delta _{r'}}{2 {\left( \left( (r')^2+a^2\right) ^2-a^2 \sin ^2\theta _{\star } \Delta _{r'}\right) ^{3/2}}} {\,\textrm{d}}r'\Biggr )\nonumber \\{} & {} \qquad \qquad \qquad \qquad \times \Delta _{\theta }\frac{(r^2+a^2)^2-a^2 \sin ^2\theta _{\star } \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r} \frac{ \left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n }{\sin ^{2n}(2 {\theta _{\star }}) }\nonumber \\{} & {} \ \ \ \qquad + \frac{a^2 \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r} \frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})} \frac{\sin (2 {\theta })}{ \sin (2 {\theta _{\star }})} \frac{ \left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n}{\sin ^{2 n}(2 {\theta _{\star }})}. \end{aligned}$$
(54)

Proof

The proof of (53) is immediate using (41), as

$$\begin{aligned}{} & {} \partial _{r_{\star }}\left( \frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}\right) \\{} & {} \qquad =-\,\frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}\frac{\partial {\theta }}{\partial r_{\star }}. \end{aligned}$$

By differentiation we obtain

$$\begin{aligned}{} & {} \partial _{\theta _{\star }}\left( \frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}\right) \nonumber \\{} & {} \qquad =-\,\frac{4 n \cos (2 {\theta _{\star }})}{\sin ^{2 n+1}(2 {\theta _{\star }})} \int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{D(\theta ',\theta _{\star })}} {\,\textrm{d}}{\theta '}\nonumber \\{} & {} \qquad \ \ \ +\frac{n}{\sin ^{2 n-1}(2 {\theta _{\star }})} \int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^{n-1}}{\sqrt{D(\theta ',\theta _{\star })}} {\,\textrm{d}}{\theta '}\nonumber \\{} & {} \qquad \ \ \ -\,\frac{1}{2 \sin ^{2 n-1}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\Delta _{\theta '} \left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{\left( D(\theta ',\theta _{\star })\right) ^3}} {\,\textrm{d}}{\theta '} \nonumber \\{} & {} \qquad \ \ \ -\,\frac{\partial {\theta }}{\partial {\theta _{\star }}}\frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n}{\sin ^{2 n}(2 {\theta _{\star }}) \sqrt{D(\theta ,\theta _{\star })}}. \end{aligned}$$
(55)

We keep the two first integrals unchanged and use (35) on the integral marked (55) to obtain

$$\begin{aligned}{} & {} -\,\frac{4 n \cos (2 \theta _{\star })}{\sin ^{2 n+1}(2 \theta _{\star })}\int _\theta ^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{D(\theta ',\theta _{\star })}} {\,\textrm{d}}\theta '\\{} & {} +\frac{n}{\sin ^{2 n-1}(2 \theta _{\star })} \int _\theta ^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^{n-1}}{\sqrt{D(\theta ',\theta _{\star })}} {\,\textrm{d}}\theta '\\{} & {} -\,\frac{1}{\sin ^{2 n+1}(2 \theta _{\star })}\int _\theta ^{\theta _{\star }} \frac{(\cos (2 \theta ')+\cos (2 \theta _{\star })) \left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{D(\theta ',\theta _{\star })}}{\,\textrm{d}}\theta '\\{} & {} -\,\frac{1}{\sin ^{2 n+1}(2 \theta _{\star })}\int _\theta ^{\theta _{\star }}\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n\sin (2 \theta ')\,\partial _{\theta '} \left( \frac{1}{\sqrt{D(\theta ',\theta _{\star })}}\right) {\,\textrm{d}}\theta '\\{} & {} -\,\frac{\partial {\theta }}{\partial {\theta _{\star }}}\frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n}{\sin ^{2 n}(2 {\theta _{\star }}) \sqrt{D(\theta ,\theta _{\star })}}. \end{aligned}$$

Integrating the last integral by parts and using the fact that

$$\begin{aligned}{} & {} -4 n\cos (2 \theta _{\star })\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) +n\sin ^2(2\theta _{\star }) - (\cos (2 \theta ')+\cos (2 \theta _{\star })) \left( \sin ^2\theta _{\star }-\sin ^2\theta \right) \\{} & {} \ \ \ +2 \cos (2 \theta ')\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) -n\sin ^2(2\theta ') \ =\ 2 (2 n + 1) (\sin ^2\theta _{\star }- \sin ^2\theta ')^2, \end{aligned}$$

we obtain

$$\begin{aligned}{} & {} \partial _{\theta _{\star }}\left( \frac{1}{\sin ^{2 n}(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^n}{\sqrt{D(\theta ',\theta _{\star })}} {\,\textrm{d}}{\theta '}\right) \nonumber \\{} & {} \qquad = \frac{2 (2 n+1) }{\sin ^{2 n+1}(2 {\theta _{\star }})}\int _{\theta '}^{\theta _{\star }} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta '\right) ^{n+1}}{\sqrt{D(\theta ',\theta _{\star })}} {\,\textrm{d}}{\theta '}\nonumber \\{} & {} \qquad \ \ \ +\left( \frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}-\,\frac{\partial {\theta }}{\partial {\theta _{\star }}}\right) \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) ^n}{\sin ^{2 n}(2 {\theta _{\star }}) \sqrt{D(\theta ,\theta _{\star })}}. \end{aligned}$$
(56)

From equality (42), it follows that

$$\begin{aligned}{} & {} \frac{\partial {\theta }}{\partial {\theta _{\star }}}-\,\frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}\nonumber \\{} & {} \ \ =\Delta _{\theta }\frac{(r^2+a^2)^2-a^2 \sin ^2\theta _{\star } \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta ' \Delta _r} \frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}\nonumber \\{} & {} \ \ \ \ \times \Biggl (2 \int _{\theta }^{\theta _{\star }} \frac{\sin ^2\theta _{\star }-\sin ^2\theta '}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '}- \int _r^{r_+} \frac{a^3 \sin ^2 (2 {\theta _{\star }}) \Delta _{r'}}{2 \left( \left( (r')^2+a^2\right) ^2-a^2 \sin ^2\theta _{\star } \Delta _{r'}\right) ^{3/2}} {\,\textrm{d}}r'\Biggr )\nonumber \\{} & {} \ \ \ \ -a^2 \Delta _r \frac{\sin ^2\theta _{\star }\Delta _{\theta } -\sin ^2\theta }{ (r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}\frac{ \sin (2 {\theta })}{\sin (2 {\theta _{\star }})}. \end{aligned}$$
(57)

Using the last equality in (56) and dividing by \(\sin (2\theta _{\star })\), we obtain (54). \(\square\)

Derivatives of \(R_2\). The following identities hold:

$$\begin{aligned}{} & {} \partial _{r_{\star }} \frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})} = -\,\frac{a}{2}\frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}\frac{\Delta _r \Delta _{\theta } \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) }{\left( (r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r\right) }, \end{aligned}$$
(58)
$$\begin{aligned}{} & {} \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}\nonumber \\{} & {} \qquad =2\frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) }{\sin ^2(2 {\theta _{\star }})}\frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}\nonumber \\{} & {} \qquad \ \ \ +\frac{1}{2} \frac{\sin ^2(2 {\theta })}{\sin ^2(2 {\theta _{\star }})} \frac{ \sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}\frac{a^2 \Delta _r\left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) }{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r} \nonumber \\{} & {} \qquad \ \ \ -\,\frac{1}{2}\frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }}) }\Delta _{\theta }\frac{((r^2+a^2)^2 -a^2 \Delta _r \sin ^2\theta _{\star })\left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) }{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}\nonumber \\{} & {} \qquad \qquad \ \ \ \times \Biggl (\frac{2}{\sin ^2 (2 {\theta _{\star }})}\int _\theta ^{\theta _{\star }} \frac{\sin ^2\theta _{\star }-\sin ^2\theta '}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '} \Biggr .\nonumber \\{} & {} \qquad \qquad \ \ \ \ \ \ \ \ \Biggl . - \int _r^{r_+} \frac{a^3 \Delta _{r'}}{2 \left( \left( (r')^2+a^2\right) ^2-a^2 \sin ^2\theta _{\star } \Delta _{r'}\right) ^{3/2}}{\,\textrm{d}}r'\Biggr ). \end{aligned}$$
(59)

Proof

From (41), we get (58). To start the proof of (59), note that

$$\begin{aligned} \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}& = -\,\frac{2 \cos (2\theta _{\star }) \sqrt{D(\theta ,\theta _{\star })}}{\sin ^3(2\theta _{\star })}\\{} & {} +\frac{1}{2} \frac{\Delta _{\theta }}{\sin (2\theta _{\star }) \sqrt{D(\theta ,\theta _{\star })}}\\{} & {} -\,\frac{1}{2}\frac{1}{\sin ^2(2\theta _{\star })}\frac{\sin (2\theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) }{\sqrt{D(\theta ,\theta _{\star })}}\frac{\partial \theta }{\partial \theta _{\star }}. \end{aligned}$$

For any A, the last expression is equal to

$$\begin{aligned}{} & {} =\frac{1}{2 \sin ^3(2\theta _{\star }) \sqrt{D(\theta ,\theta _{\star })}}\Biggl (-4 \cos (2\theta _{\star }) D(\theta ,\theta _{\star })\Biggr .\\{} & {} \ \ \ +\sin ^2(2\theta _{\star }) \Delta _{\theta }-\sin ^2(2\theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \\{} & {} \ \ \ -(A-1) \sin ^2(2\theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \\{} & {} \ \ \ \Biggl . +\sin (2\theta )\left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \sin (2\theta _{\star }) \left( A \frac{\sin (2\theta )}{\sin (2\theta _{\star })} -\,\frac{\partial \theta }{\partial \theta _{\star }} \right) \Biggr ), \end{aligned}$$

because the terms with A cancel out and the terms with \(\sin ^2(2\theta )\) cancel out. The value of A will be chosen taking (42) into account, that is we choose \(A=\Delta _{\theta }\frac{(r^2+a^2)^2 -a^2 \sin ^2\theta _{\star } \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}\). The reason we made a term with \(A-1\) appear is that such a term is proportional to \(\Delta _r\). In fact,

$$\begin{aligned} A-1=\Delta _{\theta }\frac{(r^2+a^2)^2 -a^2 \sin ^2\theta _{\star } \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}-1=-\,\frac{a^2 \Delta _r \left( \sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta \right) }{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}. \end{aligned}$$

According to (35), we have that

$$\begin{aligned}{} & {} -4 \cos (2\theta _{\star }) D(\theta ,\theta _{\star }) +\sin ^2(2\theta _{\star }) \Delta _{\theta }-\sin ^2(2\theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \\{} & {} \qquad =-4 \cos (2\theta _{\star }) D(\theta ,\theta _{\star }) +\sin ^2(2\theta _{\star }) \Delta _{\theta }+\sin (2\theta ) \partial _\theta D(\theta ,\theta _{\star })\\{} & {} \qquad =-4 \cos (2\theta _{\star }) D(\theta ,\theta _{\star }) +2(\cos (2\theta )+\cos (2\theta _{\star }))D(\theta ,\theta _{\star })\\{} & {} \qquad = 2(\cos (2\theta )-\cos (2\theta _{\star }))D(\theta ,\theta _{\star })\\{} & {} \qquad =4 \left( \sin ^2\theta _{\star }-\sin ^2\theta \right) D(\theta ,\theta _{\star }). \end{aligned}$$

Hence, the expression above is equal to

$$\begin{aligned}{} & {} \frac{1}{2 \sin ^3(2\theta _{\star }) \sqrt{D(\theta ,\theta _{\star })}}\Biggl (4 \left( \sin ^2\theta _{\star }-\sin ^2\theta \right) D(\theta ,\theta _{\star })\Biggr .\\{} & {} \ \ \ -(A-1) \sin ^2(2\theta ) \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \\{} & {} \ \ \ -\sin (2\theta )\left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) \Biggl .\sin (2\theta _{\star })\left( \frac{\partial \theta }{\partial \theta _{\star }} -A\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\right) \Biggr )\\{} & {} =2\frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) }{\sin ^2(2\theta _{\star })} \frac{\sqrt{D(\theta ,\theta _{\star })}}{\sin (2\theta _{\star })}\\{} & {} \ \ \ +\frac{1}{2}\frac{\sin ^2(2\theta )}{\sin ^2(2\theta _{\star })}\frac{\sqrt{D(\theta ,\theta _{\star })}}{\sin (2\theta _{\star })} \frac{a^2 \Delta _r\left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) }{\left( (r^2+a^2)^2 \Delta _\theta -a^2 \sin ^2\theta \Delta _r\right) }\\{} & {} \ \ \ -\,\frac{1}{2}\frac{\sin (2\theta )}{\sin ^2(2\theta _{\star })}\frac{ \left( 1+\frac{\Lambda }{3} a^2 \sin ^2\theta _{\star }\right) }{ \sqrt{D(\theta ,\theta _{\star })}}\left( \frac{\partial \theta }{\partial \theta _{\star }} -A\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\right) . \end{aligned}$$

The final expression (59) is obtained using (42). \(\square\)

Derivatives of \(R_3\). The following identities hold:

$$\begin{aligned}{} & {} \frac{\partial }{\partial {r_{\star }}}\frac{(\sin ^2\theta _{\star } -\sin ^2\theta )}{\sin ^2(2 {\theta _{\star }})}\nonumber \\{} & {} \qquad =-\,\frac{\sin (2\theta )}{\sin (2 {\theta _{\star }})} \frac{\sqrt{ \sin ^2{\theta _{\star }}\Delta _{\theta }-\sin ^2{\theta }}}{\sin (2 {\theta _{\star }})} \frac{a \Delta _r \Delta _{\theta }}{\left( (r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2{\theta }\Delta _r\right) }, \end{aligned}$$
(60)
$$\begin{aligned}{} & {} \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\frac{(\sin ^2\theta _{\star } -\sin ^2\theta )}{\sin ^2(2 {\theta _{\star }})}\\{} & {} \qquad \ \ = 4\frac{(\sin ^2\theta _{\star }-\sin ^2\theta )^2}{\sin ^4(2\theta _{\star })}\\{} & {} \qquad \ \ \ \ \ +\frac{\left( \sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta \right) }{\sin ^2(2\theta _{\star })} \frac{\sin ^2(2\theta )}{\sin ^2(2\theta _{\star })} \frac{a^2 \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}\\{} & {} \qquad \ \ \ \ \ -\,\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\frac{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2\theta _{\star })} \Delta _{\theta }\frac{(r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r}{ (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}}\\{} & {} \qquad \qquad \qquad \times \Biggl (\frac{2}{\sin ^2(2\theta _{\star })} \int _{\theta }^{\theta _{\star }} \frac{(\sin ^2\theta _{\star }-\sin ^2\theta ')}{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta '}-\sin ^2\theta '}}{\,\textrm{d}}\theta '\Biggr .\\{} & {} \qquad \qquad \qquad \ \ \ \ \Biggl .-\int _r^{r_+} \frac{a^3 \Delta _{r'}}{2\left( (r'^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _{r'}\right) ^{\frac{3}{2}}}{\,\textrm{d}}r'\Biggr ). \end{aligned}$$

Proof

From (41), we get (60). Arguing as in the proof of (59), we obtain

$$\begin{aligned} \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\frac{(\sin ^2\theta _{\star } -\sin ^2\theta )}{\sin ^2(2 {\theta _{\star }})}& = -\,\frac{4 \cos (2\theta _{\star }) (\sin ^2\theta _{\star } -\sin ^2\theta )}{\sin ^4(2\theta _{\star })} + \frac{1}{\sin ^2(2\theta _{\star })} -\,\frac{\sin (2\theta )}{\sin ^3(2\theta _{\star })}\frac{\partial \theta }{\partial \theta _{\star }}\\& = -\,\frac{4 \cos (2\theta _{\star }) (\sin ^2\theta _{\star } -\sin ^2\theta )}{\sin ^4(2\theta _{\star })} + \frac{1}{\sin ^2(2\theta _{\star })} -\,\frac{\sin ^2(2\theta )}{\sin ^4(2\theta _{\star })}\\{} & {} -(A-1)\frac{\sin ^2(2\theta )}{\sin ^4(2\theta _{\star })} -\,\frac{\sin (2\theta )}{\sin ^3(2\theta _{\star })} \Biggl (\frac{\partial \theta }{\partial \theta _{\star }}-A\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\Biggr )\\& = 4\frac{(\sin ^2\theta _{\star }-\sin ^2\theta )^2}{\sin ^4(2\theta _{\star })}\\{} & {} +\frac{\left( \sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta \right) }{\sin ^2(2\theta _{\star })} \frac{\sin ^2(2\theta )}{\sin ^2(2\theta _{\star })} \frac{a^2 \Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}\\{} & {} -\,\frac{\sin (2\theta )}{\sin ^3(2\theta _{\star })} \Biggl (\frac{\partial \theta }{\partial \theta _{\star }}-A\frac{\sin (2\theta )}{\sin (2\theta _{\star })}\Biggr ). \end{aligned}$$

One concludes by once again applying (42). \(\square\)

Derivatives of \(R_4\). The following identities hold:

$$\begin{aligned} \partial _{r_{\star }} \left( \frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}\right)& = \frac{ \sqrt{\sin ^2\theta _{\star } \Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}\frac{2 a \Delta _r \Delta _{\theta } \cos (2 {\theta })}{\left( (r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r\right) },\end{aligned}$$
(61)
$$\begin{aligned}{} & {} \left( \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\right) \left( \frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}\right) \nonumber \\{} & {} \ \ \ = 4\frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})} \frac{\left( \sin ^2\theta _{\star }-\sin ^2\theta \right) }{\sin ^2(2 {\theta _{\star }})} -2\frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})} \frac{\left( \sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta \right) }{\sin ^2(2 {\theta _{\star }})}\frac{ a^2 \Delta _r\cos (2 {\theta })}{(r^2+a^2)^2 \Delta _{\theta } -a^2 \sin ^2\theta \Delta _r}\nonumber \\{} & {} \qquad +2 \frac{\sqrt{\sin ^2\theta _{\star }\Delta _{\theta }-\sin ^2\theta }}{\sin (2 {\theta _{\star }})}\Delta _{\theta }\frac{(r^2+a^2)^2-a^2 \sin ^2\theta _{\star }\Delta _r}{(r^2+a^2)^2 \Delta _{\theta }-a^2 \sin ^2\theta \Delta _r}\cos (2 {\theta })\nonumber \\{} & {} \qquad \times \Biggl (\frac{2}{\sin ^2(2 {\theta _{\star }})}\int _{\theta }^{\theta _{\star }} \frac{\sin ^2\theta _{\star }-\sin ^2\theta '}{\sqrt{\sin ^2\theta _{\star } \Delta _{\theta '}-\sin ^2\theta '}} {\,\textrm{d}}{\theta '} - \int _r^{r_+}{ \frac{a^3 \Delta _{r'}}{2 \left( \left( (r')^2+a^2\right) ^2-a^2 \sin ^2\theta _{\star } \Delta _{r'}\right) ^{3/2}} {\,\textrm{d}}r'}\Biggr ). \end{aligned}$$
(62)

Proof

From (41), we immediately get (61). Moreover

$$\begin{aligned} \left( \frac{1}{\sin (2 {\theta _{\star }})}\partial _{\theta _{\star }}\right) \left( \frac{\sin (2 {\theta })}{\sin (2 {\theta _{\star }})}\right)& = -\,\frac{2 \sin (2 {\theta }) \cos (2 {\theta _{\star }})}{\sin ^3(2 {\theta _{\star }})}+\frac{2 \cos (2 {\theta })}{\sin ^2(2 {\theta _{\star }})}\frac{\partial {\theta }}{\partial {\theta _{\star }}}\\& = -\,\frac{\sin (2\theta )}{\sin (2\theta _{\star })} \Biggl (\frac{2\cos (2\theta _{\star })}{\sin ^2(2\theta _{\star })} -\,\frac{2\cos (2\theta )}{\sin ^2(2\theta _{\star })}\Biggr )\\{} & {} +\frac{2 \cos (2 {\theta })}{\sin ^2(2 {\theta _{\star }})}\Biggl (\frac{\partial {\theta }}{\partial {\theta _{\star }}}- \,\frac{\sin (2\theta )}{\sin (2\theta _{\star })} \Biggr ). \end{aligned}$$

Equality (62) is obtained using (57). \(\square\)

Proof of Lemma 2.11

Proof of (44). We have seen in (43) that \(\left| \frac{\partial r}{\partial {r_{\star }}}\right| \lesssim |\Delta _r|\) (this is (44) with \(k_1=1\) and \(k_2=0\)) and that \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial r}{\partial {\theta _{\star }}}\right| \lesssim |\Delta _r|\) (this is (44) with \(k_1=0\) and \(k_2=1\)). Thus, inequalities (44) follow from

  1. (i)

    (51) holds,

  2. (ii)

    (52) holds,

  3. (iii)

    \(\left| \frac{\partial r}{\partial {r_{\star }}}\right| \lesssim |\Delta _r|\), which in particular implies that \(\left| \frac{\partial \Delta _r}{\partial {r_{\star }}}\right| \lesssim |\Delta _r|\),

  4. (iv)

    \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial r}{\partial {\theta _{\star }}}\right| \lesssim |\Delta _r|\), which in particular implies that \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial \Delta _r}{\partial {\theta _{\star }}}\right| \lesssim |\Delta _r|\),

  5. (v)

    \(\left| \frac{\partial \theta }{\partial {r_{\star }}}\right| \lesssim |\Delta _r|\sin (2\theta _{\star })\lesssim |\Delta _r|\lesssim 1\),

  6. (vi)

    \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial (\sin ^2\theta )}{\partial \theta _{\star }}\right| \lesssim 1\), \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial (\sin (2\theta )\sin (2\theta _{\star }))}{\partial \theta _{\star }}\right| \lesssim 1\), \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial (\cos (2\theta ))}{\partial \theta _{\star }}\right| \lesssim 1\). \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial (\sin ^2\theta _{\star })}{\partial \theta _{\star }}\right| =1\), \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial (\sin ^2(2\theta _{\star }))}{\partial \theta _{\star }}\right| \lesssim 1\), \(\left| \frac{1}{\sin (2\theta _{\star })}\frac{\partial (\cos (2\theta _{\star }))}{\partial \theta _{\star }}\right| =2\).

\(\square\)

Proof of (45). It was shown in (43) that \(\left| \frac{\partial \theta }{\partial {r_{\star }}}\right| \lesssim |\Delta _r|\sin (2\theta _{\star })\) (this is (45) with \(k_1=1\)). Hence, inequalities (45) follow from (i), (iii) and (v) (with the bound 1). \(\square\)

Proof of (46). We have seen in (43) that \(\left| \frac{\partial \theta }{\partial {\theta _{\star }}}\right| \lesssim 1\) (this is (46) with \(k_1=0\)). Thus, inequalities (46) are a consequence of (ii), (iv) and (vi). \(\square\)

Proof of (47). Inequalities (47) result from (46), (i), (iii) and (v) (with the bound \(|\Delta _r|\)). \(\square\)

Proof of (48) and (49). These inequalities ensue from (37), (44) to (47) and (50). \(\square\)

This completes the proof of Lemma 2.11.

Remark 2.14

Note that (48) and (49) hold for any smooth function of \(r_\star\) and \(\sin ^2\theta _{\star }\).

Lemma 2.15

The functions \(\frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}h\), \(b^{\phi _{\star }}\), \({g \hspace{-4.83322pt}/}_{\theta _{\star },\theta _{\star }}\), \(\frac{{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}}{\sin ^2\theta _{\star }\sin (2\theta _\star )}\), \(\frac{{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}}{\sin ^2\theta _{\star }}\) and \(\frac{1}{\sin ^2\theta _{\star }}\left( {g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} -\,\frac{{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}}{\sin ^2(\theta _{\star })}\right)\) are smooth.

Proof

Since \(h(0,\theta _{\star })=0\), we have that \(\partial _{\theta _{\star }}h(0,\theta _{\star })=0\). So, as \(\partial _{r_{\star }}h=-B\), it follows that

$$\begin{aligned} \partial _{\theta _{\star }}h(r_{\star },\theta _{\star })& = \int _0^{r_\star } (\partial _{r_{\star }}\partial _{\theta _{\star }}h)(r',\theta _{\star }){\,\textrm{d}}r'\ =\ \int _0^{r_\star } (\partial _{\theta _{\star }}\partial _{r_{\star }}h)(r',\theta _{\star }){\,\textrm{d}}r'\\& = -\int _0^{r_\star }( (\partial _{\theta _{\star }}r)(\partial _{r}B)+ (\partial _{\theta _{\star }}\theta )(\partial _\theta B))(r',\theta _{\star }) {\,\textrm{d}}r'\\& = -\sin (2\theta _{\star })\int _0^{r_\star }\Biggl (\left( \frac{1}{\sin (2\theta _{\star })} \partial _{\theta _{\star }}r\right) (\partial _{r}B)\Biggr .\\{} & {} \qquad \qquad \qquad +\Biggl . (\partial _{\theta _{\star }}\theta )\left( \frac{1}{\sin (2\theta )}\partial _\theta B\right) \frac{\sin (2\theta )}{\sin (2\theta _{\star })}\Biggr )(r',\theta _{\star }) {\,\textrm{d}}r'. \end{aligned}$$

Recall from (40) that \(\partial _{\theta _{\star }}r\) has a factor \(\Delta _r\sin (2\theta _{\star })\) and observe that

$$\begin{aligned} \frac{1}{\sin (2\theta )} \partial _\theta B=\frac{\Xi a^3\Delta _r((r^2+a^2) \left( 1-\,\frac{\Lambda }{3}r^2\right) -\Delta _r)}{\Upsilon ^2}. \end{aligned}$$

This shows that \(\frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}h\) has derivatives of all orders with respect to \(\partial _{r_{\star }}\) and \(\frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\). Moreover, as \(b^{\phi _{\star }}=2B\), we obtain

$$\begin{aligned} \left| \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}b^{\phi _{\star }} \right| \lesssim |\Delta _r| \end{aligned}$$

and \({g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }}\) is smooth.

Note that \(\frac{\sin ^2\theta }{\sin ^2\theta _{\star }}\) has derivatives of all orders with respect to \(r_\star\) and \(\sin ^2\theta _{\star }\). Indeed, we have

$$\begin{aligned} \partial _{r_{\star }} \left( \frac{\sin ^2\theta }{\sin ^2\theta _{\star }}\right)& = \frac{\sin (2\theta )}{\sin ^2\theta _{\star }}\frac{\partial \theta }{\partial r_\star }. \end{aligned}$$

As \(\frac{\partial \theta }{\partial r_\star }\) has a factor \(\sin (2\theta _\star )\), the right-hand side has a factor \(\sin (2\theta )\sin (2\theta _\star )\). Taking (50) into account, differentiability at \(\theta _{\star }\) equal to \(\frac{\pi }{2}\) is not a problem. Neither is there a problem at \(\theta _{\star }\) equal to zero because

$$\begin{aligned} \frac{\sin (2\theta )\sin (2\theta _{\star })}{\sin ^2\theta _{\star }}= 4\frac{\sin (2\theta )}{\sin (2\theta _{\star })}(1-\sin ^2\theta _\star ). \end{aligned}$$

Differentiability with respect to \(\sin ^2\theta _{\star }\) holds at \(\theta _{\star }\) equal to zero because

$$\begin{aligned} \frac{\sin ^2\theta }{\sin ^2\theta _{\star }}= \left( \frac{\sin (2\theta )}{\sin (2\theta _{\star })}\right) ^2 \frac{1-\sin ^2\theta _{\star }}{1-\sin ^2\theta }, \end{aligned}$$

and differentiability holds at \(\theta _{\star }\) equal to \(\frac{\pi }{2}\) because

$$\begin{aligned} \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }} \left( \frac{\sin ^2\theta }{\sin ^2\theta _{\star }}\right)& = \frac{\sin (2\theta )}{\sin (2\theta _{\star })} \frac{\partial \theta }{\partial \theta _\star } \frac{1}{\sin ^2\theta _{\star }}-\,\frac{\sin ^2\theta }{\sin ^2\theta _{\star }} \frac{1}{\sin ^2\theta _{\star }}. \end{aligned}$$

So, we see that

$$\begin{aligned} \frac{{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}}{\sin ^2\theta _{\star }\sin (2\theta _\star )}\qquad \hbox {and}\qquad \frac{{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}}{\sin ^2\theta _{\star }} \end{aligned}$$

are smooth. Using (13) and (37), we get

$$\begin{aligned}{} & {} \frac{1}{\sin ^2\theta _{\star }}\left( {g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} -\,\frac{{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}}{\sin ^2(\theta _{\star })}\right) \nonumber \\{} & {} = \frac{1}{\sin ^2\theta _{\star }}\left( \frac{\rho ^2L^2}{\Upsilon } -\,\frac{\sin ^2\theta }{\sin ^2\theta _{\star }}\frac{\Upsilon }{\Xi ^2\rho ^2}\right) +\sin ^2(2\theta _{\star })\frac{\sin ^2\theta }{\sin ^2\theta _{\star }} \frac{\Upsilon }{\Xi ^2\rho ^2} \left( \frac{\partial _{\theta _{\star }}h}{\sin (2\theta _{\star })}\right) ^2\nonumber \\{} & {} = \frac{1}{\Upsilon \rho ^2\sin ^2\theta _{\star }}\left( {\rho ^4L^2} -\,\frac{\sin ^2\theta }{\sin ^2\theta _{\star }}\frac{\Upsilon ^2}{\Xi ^2}\right) +\hbox {smooth}\nonumber \\{} & {} = \frac{1}{\Xi ^2\Upsilon \rho ^2\sin ^2\theta _{\star }} \left( (r^2+a^2\cos ^2\theta )^2\Xi ^2((r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r) \frac{\sin ^2(2\theta )}{\sin ^2(2\theta _{\star })}\right. \nonumber \\{} & {} \qquad \qquad \qquad \ \ \left. -\,\frac{\sin ^2\theta }{\sin ^2\theta _{\star }}((r^2+a^2)^2\Delta _\theta -a^2\sin ^2\theta \Delta _r)^2 \right) +\hbox {smooth}\nonumber \\{} & {} = \frac{1}{\Xi ^2\Upsilon \rho ^2\sin ^2\theta _{\star }}\frac{\sin ^2\theta }{\sin ^2\theta _{\star }} \frac{1}{\cos ^2\theta _{\star }} \left[ (r^2+a^2\cos ^2\theta )^2\Xi ^2((r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r) \cos ^2\theta \right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \qquad \ \ \left. -((r^2+a^2)^2\Delta _\theta -a^2\sin ^2\theta \Delta _r)^2\cos ^2\theta _{\star } \right] +\hbox {smooth}.\nonumber \\{} & {} = \frac{1}{\Xi ^2\Upsilon \rho ^2}\frac{\sin ^2\theta }{\sin ^2\theta _{\star }} \frac{1}{1-\sin ^2\theta _{\star }}\frac{p(\sin ^2\theta ,\sin ^2\theta _{\star })}{\sin ^2\theta _{\star }}+\hbox {smooth}. \end{aligned}$$
(63)

The expression inside the square parenthesis is a polynomial p in \(\alpha :=\sin ^2\theta\) and \(\beta :=\sin ^2\theta _{\star }\) satisfying \(p(0,0)=0\). So, \(p(\alpha ,\beta )=D_1p(0,0)\alpha +D_2p(0,0)\beta +\) higher order terms. Thus, it is clear that \(\frac{p(\sin ^2\theta ,\sin ^2\theta _{\star })}{\sin ^2\theta _{\star }}\) is smooth, and so it is clear that (63) is smooth at zero. Differentiability at \(\frac{\pi }{2}\) is not an issue since

$$\begin{aligned} \frac{1}{\sin (2\theta _{\star })}\partial _{\theta _{\star }}\frac{1}{\sin ^2\theta _{\star }} =-\,\frac{1}{\sin ^4\theta _{\star }}. \end{aligned}$$

\(\square\)

Remark 2.16

The Christoffel symbols of the metric \({g \hspace{-4.83322pt}/}\) are bounded with the exception of \({\Gamma \hspace{-6.66656pt}/ \,}^{\phi _{\star }}_{\theta _{\star }\phi _{\star }}\) which blows up precisely like \(\frac{1}{\sin \theta _{\star }}\).

Proof

Because \({g \hspace{-4.83322pt}/}\) behaves like the round metric on \({{\mathbb {S}}}^2\), \({g \hspace{-4.83322pt}/}^{-1}\) behaves like the inverse of the round metric on \({{\mathbb {S}}}^2\), in that \({g \hspace{-4.83322pt}/}^{\phi _{\star }\phi _{\star }}\) blows up at \(\theta _{\star }=0\) like \(\frac{1}{\sin ^2\theta _{\star }}\). Hence, the Christoffel symbols of the metric \({g \hspace{-4.83322pt}/}\) also behave like the ones of the round metric on \({{\mathbb {S}}}^2\), namely, they are all smooth with the exception of \({\Gamma \hspace{-6.66656pt}/ \,}^{\phi _{\star }}_{\theta _{\star }\phi _{\star }}\),

$$\begin{aligned} {\Gamma \hspace{-6.66656pt}/ \,}^{\phi _{\star }}_{\theta _{\star }\phi _{\star }}& = \frac{1}{2} {g \hspace{-4.83322pt}/}^{\phi _{\star }\alpha }(\partial _{\theta _{\star }}{g \hspace{-4.83322pt}/}_{\alpha \phi _{\star }} +\partial _{\phi _{\star }}{g \hspace{-4.83322pt}/}_{\theta _{\star }\alpha }-\partial _\alpha {g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }})\\& = \hbox {smooth}+\frac{1}{2} {g \hspace{-4.83322pt}/}^{\phi _{\star }{\phi _{\star }}}(\partial _{\theta _{\star }}{g \hspace{-4.83322pt}/}_{{\phi _{\star }}\phi _{\star }} +\partial _{\phi _{\star }}{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}-\partial _{\phi _{\star }}{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }})\\& = \frac{1}{2} {g \hspace{-4.83322pt}/}^{\phi _{\star }{\phi _{\star }}}\partial _{\theta _{\star }}{g \hspace{-4.83322pt}/}_{{\phi _{\star }}\phi _{\star }}+\hbox {smooth}\\& = \frac{1}{2}\frac{\Xi ^2 \rho ^2}{\Upsilon \sin ^2\theta } \partial _{\theta _{\star }} \left( \frac{\sin ^2\theta }{\Xi ^2\rho ^2}\Upsilon \right) +\hbox {smooth}\\& = \frac{1}{2}\frac{1}{\sin ^2\theta }\partial _{\theta _{\star }}(\sin ^2\theta )+\hbox {smooth}\\& = \frac{\cos \theta }{\sin \theta }\frac{\partial \theta }{\partial \theta _{\star }}+\hbox {smooth}\\& = \Delta _{\theta }\frac{(r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r}{ (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}}{\frac{\sin (2\theta )}{\sin (2\theta _{\star })}}\frac{\cos \theta }{\sin \theta }+\hbox {smooth}\\& = \Delta _{\theta }\frac{(r^2+a^2)^2-a^2\sin ^2\theta _{\star }\Delta _r}{ (r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _{r}}{\frac{\cos ^2\theta }{\cos \theta _{\star }}}\frac{1}{\sin \theta _{\star }}+\hbox {smooth}, \end{aligned}$$

which blows up precisely like \(\frac{1}{\sin \theta _{\star }}\). \(\square\)

2.2.6 Regularity at \(\theta _\star =0\) and \(\theta _\star =\pi\)

Note that the coordinates that are constructed are not just locally defined but are well-defined globally on a manifold diffeomorphic to \(\mathbb {R}^2\times \mathbb {S}^2\).

Regularity of the metric.

We check the regularity of the metric at \(\theta _\star =0\) using coordinates

$$\begin{aligned} \begin{array}{rcl} x&{}=&{}\sin \theta _\star \cos \phi _\star ,\\ y&{}=&{}\sin \theta _\star \sin \phi _\star . \end{array} \end{aligned}$$
(64)

Lemma 2.17

The metric is smooth at \(\theta _\star =0\) and

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}g= -2\Omega ^2 \left( {\textrm{d}}u\otimes {\textrm{d}}v+{\,\textrm{d}}v\otimes {\textrm{d}}u\right) + \frac{r^2+a^2}{\Xi }\left( \frac{\partial \theta }{\partial \theta _\star }\right) ^2(r_\star ,0)\left( {\,\textrm{d}}x\otimes {\,\textrm{d}}x+{\,\textrm{d}}y\otimes {\,\textrm{d}}y\right) . \end{aligned}$$
(65)

Proof

The relations

$$\begin{aligned} {\,\textrm{d}}\theta _\star& = \ \ \, \frac{\cos \phi _\star }{\cos \theta _\star }{\,\textrm{d}}x +\frac{\sin \phi _\star }{\cos \theta _\star }{\,\textrm{d}}y,\\ {\,\textrm{d}}\phi _\star& = -\,\frac{\sin \phi _\star }{\sin \theta _\star }{\,\textrm{d}}x +\frac{\cos \phi _\star }{\sin \theta _\star }{\,\textrm{d}}y \end{aligned}$$

imply that

$$\begin{aligned}{} & {} \,{g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} {\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\theta _{\star } +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\phi _{\star } +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\textrm{d}}\phi _{\star }\otimes {\textrm{d}}\theta _{\star }+{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }} {\textrm{d}}\phi _{\star }\otimes {\textrm{d}}\phi _{\star }\\{} & {} \ \ \ ={g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }}\left( \frac{\cos ^2\phi _\star }{\cos ^2\theta _\star }{\,\textrm{d}}x\otimes {\,\textrm{d}}x +\frac{\sin (2\phi _\star )}{2\cos ^2\theta _\star }{\,\textrm{d}}x\otimes {\,\textrm{d}}y +\frac{\sin (2\phi _\star )}{2\cos ^2\theta _\star }{\,\textrm{d}}y\otimes {\,\textrm{d}}x +\frac{\sin ^2\phi _\star }{\cos ^2\theta _\star }{\,\textrm{d}}y\otimes {\,\textrm{d}}y \right) \\{} & {} \qquad +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}\left( -2\frac{\sin (2\phi _\star )}{\sin (2\theta _\star )}{\,\textrm{d}}x\otimes {\,\textrm{d}}x +\frac{\cos (2\phi _\star )}{\sin (2\theta _\star )}{\,\textrm{d}}x\otimes {\,\textrm{d}}y +\frac{\cos (2\phi _\star )}{\sin (2\theta _\star )}{\,\textrm{d}}y\otimes {\,\textrm{d}}x +2\frac{\sin ^2\phi _\star }{\sin (2\theta _\star )}{\,\textrm{d}}y\otimes {\,\textrm{d}}y \right) \\{} & {} \qquad +{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}\left( \frac{\sin ^2\phi _\star }{\sin ^2\theta _\star }{\,\textrm{d}}x\otimes {\,\textrm{d}}x -\,\frac{\sin (2\phi _\star )}{2\sin ^2\theta _\star }{\,\textrm{d}}x\otimes {\,\textrm{d}}y -\,\frac{\sin (2\phi _\star )}{2\sin ^2\theta _\star }{\,\textrm{d}}y\otimes {\,\textrm{d}}x +\frac{\cos ^2\phi _\star }{\sin ^2\theta _\star }{\,\textrm{d}}y\otimes {\,\textrm{d}}y \right) . \end{aligned}$$

Using (13), (22), (24) and (37), we have

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}{g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }}& = \frac{r^2+a^2}{\Xi }\left( \frac{\partial \theta }{\partial \theta _\star }\right) ^2(r_\star ,0),\\ \lim _{\theta _\star \rightarrow 0}\frac{{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }}}{\sin ^2\theta _\star }& = \frac{r^2+a^2}{\Xi }\left( \frac{\partial \theta }{\partial \theta _\star }\right) ^2(r_\star ,0). \end{aligned}$$

As, by (23),

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}\frac{{g \hspace{-4.83322pt}/}_{\theta _\star \phi _\star }}{\sin (2\theta _\star )}=0, \end{aligned}$$

we obtain

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}{g \hspace{-4.83322pt}/}=\frac{r^2+a^2}{\Xi }\left( \frac{\partial \theta }{\partial \theta _\star }\right) ^2(r_\star ,0)\left( {\,\textrm{d}}x\otimes {\,\textrm{d}}x+{\,\textrm{d}}y\otimes {\,\textrm{d}}y\right) . \end{aligned}$$

The trigonometric functions of \(\phi _\star\), which would make the metric discontinuous at \(\theta _\star =0\), have disappeared. Moreover, since

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}b^{\phi _{\star }}{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}\theta _{\star }\otimes {\,\textrm{d}}v=0, \qquad \lim _{\theta _\star \rightarrow 0}b^{\phi _{\star }}{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}v\otimes {\,\textrm{d}}\theta _{\star }=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{\theta _\star \rightarrow 0}{g \hspace{-4.83322pt}/}_{\phi _\star \phi _\star }=0, \end{aligned}$$

we conclude that (65) holds. This shows that the metric is continuous at \(\theta _\star =0\). Lemma 2.15 implies that the extension of the metric is smooth. \(\square\)

Calculation of \(\partial _{\theta _\star }\theta (r_\star ,0)\).

Lemma 2.18

We have that

$$\begin{aligned} \frac{\partial \theta }{\partial \theta _\star }(r_\star ,0)= \frac{1}{\frac{\partial \theta _\star }{\partial \theta }(r,0)}= \sqrt{\Xi }\sin \left( \arcsin \left( \frac{1}{\sqrt{\Xi }}\right) + \arctan \left( \frac{r}{a}\right) -\,\arctan \left( \frac{r_+}{a}\right) \right) . \end{aligned}$$
(66)

Proof

Using the definition of F in (9), the equation \(F(r,\theta ,\theta _\star (r,\theta ))=0\) can be written as

$$\begin{aligned} \int _\theta ^{\theta _\star }\frac{{\,\textrm{d}}\theta '}{P(\theta ',\theta _\star )} =\int _r^{r_+}\frac{{\,\textrm{d}}r'}{Q(r',\theta _\star )}. \end{aligned}$$
(67)

We will take the limit of both sides of (67) as \(\theta\) goes to 0. As \(\lim _{\theta \rightarrow 0}\theta _\star (r,\theta )=0\), uniformly in r,

$$\begin{aligned} \lim _{\theta \rightarrow 0}\int _r^{r_+}\frac{{\,\textrm{d}}r'}{Q(r',\theta _\star )}= \int _r^{r_+}\frac{{\,\textrm{d}}r'}{(r')^2+a^2}=\frac{1}{a}\arctan \left( \frac{r_+}{a}\right) - \frac{1}{a}\arctan \left( \frac{r}{a}\right) . \end{aligned}$$

Using the substitution \(s=\frac{\sin \theta '}{\sin \theta _\star }\),

$$\begin{aligned} \int _\theta ^{\theta _\star }\frac{{\,\textrm{d}}\theta '}{P(\theta ',\theta _\star )}& = \frac{1}{a} \int _{\frac{\sin \theta }{\sin \theta _\star }}^1 \frac{1}{\cos \theta '}\frac{{\,\textrm{d}}s}{\sqrt{\left( 1+\frac{\Lambda }{3}a^2\cos ^2\theta '\right) -s^2}}, \end{aligned}$$

with \(\cos \theta '=\sqrt{1-s^2\sin ^2\theta _\star }\). The last integrand converges uniformly to

$$\begin{aligned} \frac{1}{\sqrt{\Xi -s^2}} \end{aligned}$$

as \(\theta\) goes to zero. Thus, we get

$$\begin{aligned} \lim _{\theta \rightarrow 0} \int _\theta ^{\theta _\star }\frac{{\,\textrm{d}}\theta '}{P(\theta ',\theta _\star )}& = \frac{1}{a}\int _{1\big /\left( \frac{\partial \theta _\star }{\partial \theta }(r,0)\right) }^1 \frac{{\,\textrm{d}}s}{\sqrt{\Xi -s^2}}=\frac{1}{a} \arcsin \left( \frac{s}{\sqrt{\Xi }}\right) \Bigg |_{1\big /\left( \frac{\partial \theta _\star }{\partial \theta }(r,0)\right) }^1\\& = \frac{1}{a}\arcsin \left( \frac{1}{\sqrt{\Xi }}\right) -\frac{1}{a} \arcsin \left( \frac{1}{\sqrt{\Xi }\frac{\partial \theta _\star }{\partial \theta }(r,0)}\right) . \end{aligned}$$

So, from (67) we conclude that

$$\begin{aligned} \arcsin \left( \frac{1}{\sqrt{\Xi }\frac{\partial \theta _\star }{\partial \theta }(r,0)}\right) =\arcsin \left( \frac{1}{\sqrt{\Xi }}\right) + \arctan \left( \frac{r}{a}\right) -\,\arctan \left( \frac{r_+}{a}\right) . \end{aligned}$$

We know that the right-hand side of this equality is positive because we guaranteed that (32) holds. And the right-hand side is obviously smaller than \(\frac{\pi }{2}\). Therefore, we have

$$\begin{aligned} \frac{\partial \theta _\star }{\partial \theta }(r,0)= \frac{1}{\sqrt{\Xi }}\csc \left( \arcsin \left( \frac{1}{\sqrt{\Xi }}\right) + \arctan \left( \frac{r}{a}\right) -\,\arctan \left( \frac{r_+}{a}\right) \right) . \end{aligned}$$

This is strictly greater than 1 (and goes to 1 as \(r\nearrow r_+\)). According to (6) and (15), the derivative of the map \((r,\theta )\mapsto (r_\star ,\theta _\star )\) at (r, 0) is represented by the matrix

$$\begin{aligned} \left[ \begin{array}{ll} \frac{r^2+a^2}{\Delta _r}&{}0\\ 0&{}\frac{\partial \theta _\star }{\partial \theta }(r,0) \end{array} \right] . \end{aligned}$$

Equality (66) follows. \(\square\)

Regularity of functions at the poles. Using the change of coordinates (64), the variable \(\theta _\star\) is written in terms of x and y as

$$\begin{aligned} \theta _\star =\arcsin \sqrt{x^2+y^2}. \end{aligned}$$

Given a function f that transforms pairs \((r_\star ,\theta _\star )\), we want to study the differentiability of

$$\begin{aligned} (r_\star ,x,y) {\mathop { \longmapsto }\limits ^{\hat{f}}} f\left( r_\star ,\arcsin \sqrt{x^2+y^2}\right) . \end{aligned}$$

Remark 2.19

Let \(f:(-\infty ,+\infty )\times \left[ 0,\pi \right] \rightarrow {{\mathbb {R}}}\) be \(C^1\) such that \(\frac{1}{\sin (2\theta _\star )}\left( \partial _{\theta _\star }f\right)\) has a limit when \(\theta _\star =0\). Then \(\hat{f}\) is \(C^1\).

Proof

The derivative of \(\hat{f}\) with respect to x is

$$\begin{aligned} \partial _x\hat{f}& = \left( \partial _{\theta _\star }f\right) \frac{1}{\sqrt{1-(x^2+y^2)}}\frac{x}{\sqrt{x^2+y^2}}. \end{aligned}$$
(68)

We observe that, although the map \((x,y)\mapsto \sqrt{x^2+y^2}\) is not differentiable at the origin, when \(x=y=0\) the quotient \(\frac{x}{\sqrt{x^2+y^2}}\) in (68) is \(+1\) or \(-1\), according to the calculation of a right or a left derivative. If we assume that \(\frac{1}{\sin (2\theta _\star )}\left( \partial _{\theta _\star }f\right)\) has a limit at \(((r_\star )_0,0)\), then \(\partial _{\theta _\star }f((r_\star )_0,0)=0\), so there is no indetermination in (68), and

$$\begin{aligned} \partial _x\hat{f}(r_\star ,0,0)=0. \end{aligned}$$

The function \(\hat{f}\) has continuous partial derivatives with respect to \(r_\star\), x and y in a neighborhood of each point \(((r_\star )_0,0,0)\), and so it is differentiable. \(\square\)

Note that we can write (68) as

$$\begin{aligned} \partial _x\hat{f} \ =\ \left( \partial _{\theta _\star }f\right) \frac{1}{\cos \theta _\star }\frac{x}{\sin \theta _\star } \ =\ 2x\frac{1}{\sin (2\theta _\star )}\left( \partial _{\theta _\star }f\right) . \end{aligned}$$
(69)

(This is the derivative of f with respect to \(\sin ^2\theta _\star =x^2+y^2\) multiplied by the derivative of \(\sin ^2\theta _\star\) with respect to x.) Expression (69) shows that if the quotient \(\frac{1}{\sin (2\theta _\star )}\left( \partial _{\theta _\star }f\right)\) were unbounded, then one might have problems with the differentiability when \((x,y)=(0,0)\). For example, if this quotient were to behave like \(\frac{1}{\sin \theta _\star }\) around \(((r_\star )_0,0)\), then \(\partial _x\hat{f}\) would behave like \(\cos \phi _\star\), so that \(\partial _x\hat{f}((r_\star )_0,0,0)\) would not exist.

Remark 2.20

Let \(f:(-\infty ,+\infty )\times \left[ 0,\pi \right] \rightarrow {{\mathbb {R}}}\) be \(C^\infty\) in the sense of Remark 2.12. Then \(\hat{f}\) is \(C^\infty\).

Proof

It is clear that \(\partial _{r_\star }^2\hat{f}\), \(\partial _x\partial _{r_\star }\hat{f}\) and \(\partial _y\partial _{r_\star }\hat{f}\) are continuous. The continuity of \(\partial _x^2\hat{f}\) is a consequence of

$$\begin{aligned} \partial _x^2\hat{f}= \partial _x\left( 2x\frac{1}{\sin (2\theta _\star )}\left( \partial _{\theta _\star }f\right) \right) =2\frac{1}{\sin (2\theta _\star )}\left( \partial _{\theta _\star }f\right) +4x^2\left( \frac{1}{\sin (2\theta _\star )}\partial _{\theta _\star } \right) ^2f. \end{aligned}$$

The continuity of \(\partial _x\partial _y\hat{f}\) and \(\partial _y^2\hat{f}\) follow in the same way. So \(\hat{f}\) is \(C^2\). One proves by induction that \(\hat{f}\) is \(C^\infty\). \(\square\)

Remark 2.21

A differentiable function such that \(f(r_\star ,\theta _\star ) =f(r_\star ,\pi -\theta _\star )\) satisfies \(\partial _{\theta _\star }f\left( r_\star ,\frac{\pi }{2}\right) =0\). The quotient \(\frac{1}{\sin (2\theta _\star )}\partial _{\theta _\star }f\) has a finite limit at \(\left( r_\star ,\frac{\pi }{2}\right)\), provided that the numerator is analytic.

Indeed, both the numerator and the denominator vanish at \(\left( r_\star ,\frac{\pi }{2}\right)\) and the denominator has a first order zero there.

Remark 2.22

Suppose that \(f:(-\infty ,+\infty )\times [0,\pi ]\times S^1\rightarrow {{\mathbb {R}}}\) is smooth and define

$$\begin{aligned} \hat{f}(r_\star ,x,y)=f\left( r_\star ,\arcsin \sqrt{x^2+y^2},\arg (x+iy)\right) , \end{aligned}$$

where \(\phi _\star =\arg (x+iy)=\arctan \frac{y}{x}\) for \(x>0\), and otherwise \(\arg (x+iy)\) is \(\arctan \frac{y}{x}\) with an appropriate constant added. Then

$$\begin{aligned} \partial _x\hat{f}& = \frac{x}{\sqrt{1-(x^2+y^2)}\sqrt{x^2+y^2}}\partial _{\theta _\star }f -\,\frac{y}{x^2+y^2}\partial _{\phi _\star }f\\& = \frac{\cos \phi _\star }{\cos \theta _\star }\partial _{\theta _\star }f-\, \frac{\sin \phi _\star }{\sin \theta _\star }\partial _{\phi _\star }f,\\ \partial _y\hat{f}& = \frac{y}{\sqrt{1-(x^2+y^2)}\sqrt{x^2+y^2}}\partial _{\theta _\star }f +\frac{x}{x^2+y^2}\partial _{\phi _\star }f\\& = \frac{\sin \phi _\star }{\cos \theta _\star }\partial _{\theta _\star }f+ \frac{\cos \phi _\star }{\sin \theta _\star }\partial _{\phi _\star }f. \end{aligned}$$

Regularity of the change of coordinates \((t,r,\theta ,\phi )\mapsto (t,r_\star ,\theta _\star ,\phi )\). Define

$$\begin{aligned} \begin{array}{rcl} \hat{x}&{}=&{}\sin \hat{\theta }_\star \cos {\phi },\\ \hat{y}&{}=&{}\sin \hat{\theta }_\star \sin {\phi }, \end{array}\qquad \qquad \begin{array}{rcl} \tilde{x}&{}=&{}\sin \theta \cos \phi ,\\ \tilde{y}&{}=&{}\sin \theta \sin \phi . \end{array} \end{aligned}$$
  1. (i)

    The map \((t,r_\star ,\hat{x},\hat{y})\mapsto (t,r,\tilde{x},\tilde{y})\) is smooth. The derivative \(\partial _{r_\star }r\) is given by (39). Moreover, using (69), we get

    $$\begin{aligned} \partial _{\hat{x}} r& = 2{\hat{x}} \frac{1}{\sin (2\hat{\theta }_\star )} \partial _{\theta _\star }r\quad (\hbox {see}~(40)\ \hbox {for}\ \partial _{\theta _\star }r),\\ \partial _{\hat{y}} r& = 2{\hat{y}} \frac{1}{\sin (2\hat{\theta }_\star )} \partial _{\theta _\star }r,\\ \partial _{r_\star }\sin \theta& = \cos \theta \frac{\partial \theta }{\partial r_\star }\ =\ 2\frac{\sin (2\hat{\theta }_\star )}{\sin (2\theta )}\sin \theta \cos ^2\theta \left( \frac{1}{\sin (2\hat{\theta }_\star )}\partial _{r_\star }\theta \right) \quad (\hbox {see}~(41)\ \hbox {for}\ \partial _{r_\star }\theta ). \end{aligned}$$

    Applying Remark 2.22, we obtain

    $$\begin{aligned} \partial _{\hat{x}}\tilde{x}& = \frac{\cos \theta }{\cos \theta _\star }\cos ^2\phi \, \partial _{\theta _\star }\theta +\frac{\sin \theta }{\sin \theta _\star }\sin ^2\phi ,\\ \partial _{\hat{y}}\tilde{x}& = \frac{\cos \theta }{\cos \theta _\star }\sin \phi \cos \phi \, \partial _{\theta _\star }\theta -\, \frac{\sin \theta }{\sin \theta _\star }\sin \phi \cos \phi ,\\ \partial _{\hat{x}}\tilde{y}& = \frac{\cos \theta }{\cos \theta _\star }\sin \phi \cos \phi \, \partial _{\theta _\star }\theta -\, \frac{\sin \theta }{\sin \theta _\star }\sin \phi \cos \phi ,\\ \partial _{\hat{y}}\tilde{y}& = \frac{\cos \theta }{\cos \theta _\star }\sin ^2\phi \, \partial _{\theta _\star }\theta + \frac{\sin \theta }{\sin \theta _\star }\cos ^2\phi . \end{aligned}$$

    Notice that when \(\theta =0\), we have

    $$\begin{aligned} \partial _{\hat{x}}\tilde{x}& = \partial _{\theta _\star }\theta ,\\ \partial _{\hat{y}}\tilde{x}& = 0,\\ \partial _{\hat{x}}\tilde{y}& = 0,\\ \partial _{\hat{y}}\tilde{y}& = \partial _{\theta _\star }\theta . \end{aligned}$$

    The quotient \(\frac{\cos \theta }{\cos \theta _\star }\) is smooth because

    $$\begin{aligned} \frac{1}{\sin (2\theta _\star )}\partial _{\theta _\star }\left( \frac{\cos \theta }{\cos \theta _\star }\right) = -\,\frac{1}{2}\frac{\sin (2\theta )}{\sin (2\theta _\star )} \frac{\cos \theta _\star }{\cos \theta }\frac{\partial \theta }{\partial \theta _\star } +\frac{1}{2}\frac{\cos \theta }{\cos \theta _\star }\frac{1}{\cos ^2\theta _\star }. \end{aligned}$$
    (70)

    The quotient \(\frac{\sin \theta }{\sin \theta _\star }\) is smooth because

    $$\begin{aligned} \frac{\sin \theta }{\sin \theta _\star }=\frac{\sin (2\theta )}{\sin (2\theta _\star )} \frac{\cos \theta _\star }{\cos \theta }. \end{aligned}$$

    Thus r, \(\tilde{x}\) and \(\tilde{y}\) are \(C^\infty\) functions of \(r_\star\), \(\hat{x}\) and \(\hat{y}\).

  2. (ii)

    The map \((t,r,\tilde{x},\tilde{y})\mapsto (t,r_\star ,\hat{x},\hat{y})\) is smooth. Recall that \(\partial _rr_\star\) is given in (6), and

    $$\begin{aligned} \partial _r\sin \theta _\star =\cos \theta _\star \partial _r\theta _\star =\cos \theta _\star \frac{1}{GQ}= 2\frac{\sin (2\theta )}{\sin (2\theta _\star )}\sin \theta _\star \cos ^2\theta _\star \left( \frac{1}{\sin (2\theta )}\frac{1}{GQ} \right) , \end{aligned}$$

    as \(\partial _r\theta _\star\) is given in (15). Inequalities (38) yield

    $$\begin{aligned} \lim _{\theta \rightarrow 0}\frac{1}{G(r,\theta ,\theta _\star (r,\theta ))}=0. \end{aligned}$$

    One other consequence of (6) and (15) is

    $$\begin{aligned} \partial _{\tilde{x}}r_\star& = 2\tilde{x}\frac{1}{\sin (2\theta )}\partial _\theta r_\star \ =\ 2a\tilde{x}\frac{\sqrt{\sin ^2\theta _\star \Delta _\theta -\sin ^2\theta }}{\Delta _\theta \sin (2\theta )},\\ \partial _{\tilde{y}}r_\star& = 2\tilde{y}\frac{1}{\sin (2\theta )}\partial _\theta r_\star \ =\ 2a\tilde{y}\frac{\sqrt{\sin ^2\theta _\star \Delta _\theta -\sin ^2\theta }}{\Delta _\theta \sin (2\theta )}. \end{aligned}$$

    The formulas for \(\partial _{\tilde{x}}\hat{x}\), \(\partial _{\tilde{y}}\hat{x}\), \(\partial _{\tilde{x}}\hat{y}\) and \(\partial _{\tilde{y}}\hat{y}\), are similar to the ones for \(\partial _{\hat{x}}\tilde{x}\), \(\partial _{\hat{y}}\tilde{x}\), \(\partial _{\hat{x}}\tilde{y}\) and \(\partial _{\hat{y}}\tilde{y}\) (interchange \(\theta\) and \(\theta _\star\)). Recall that \(\partial _\theta \theta _\star =-\,\frac{1}{GP}\). It follows that \(r_\star\), \(\hat{x}\) and \(\hat{y}\) are \(C^\infty\) functions of r, \(\tilde{x}\) and \(\tilde{y}\).

Regularity of the change of coordinates \((t,r_\star ,\theta _\star ,\phi )\mapsto (t,r_\star ,\theta _\star ,\phi _\star )\). Recall that

$$\begin{aligned} \phi _\star =\phi -h(r_\star ,\theta _\star ). \end{aligned}$$

The fact that both \((t,r_\star ,\theta _\star ,\phi )\mapsto (t,r_\star ,\theta _\star ,\phi _\star )\) and \((t,r_\star ,\theta _\star ,\phi _\star )\mapsto (t,r_\star ,\theta _\star ,\phi )\) are \(C^\infty\) is a simple consequence of (18), which gives \(\partial _{r_\star }h\), and Lemma 2.15, which gives \(\frac{1}{\sin (2\theta _\star )}\partial _{\theta _\star }h\).

We remark that

$$\begin{aligned} \partial _x& = \frac{\cos \phi _\star }{\cos \theta _\star }\partial _{\theta _\star } -\,\frac{\sin \phi _\star }{\sin \theta _\star }\partial _{\phi _\star },\end{aligned}$$
(71)
$$\begin{aligned} \partial _y& = \frac{\sin \phi _\star }{\cos \theta _\star }\partial _{\theta _\star } +\frac{\cos \phi _\star }{\sin \theta _\star }\partial _{\phi _\star }. \end{aligned}$$
(72)

Hence, the differentiability of \((t,r_\star ,x,y)\mapsto (t,r_\star ,\hat{x},\hat{y})\) follows from

$$\begin{aligned} \partial _{r_\star }\hat{x}& = -\sin \theta _\star \sin \phi \,\partial _{r_\star }h,\\ \partial _x\hat{x}& = \cos (h(r_\star ,\theta _\star )) -\,\frac{\sin \theta _\star }{\cos \theta _\star }\cos \phi _\star \sin \phi \, \partial _{\theta _\star }h,\\ \partial _y\hat{x}& = -\sin (h(r_\star ,\theta _\star )) -\,\frac{\sin \theta _\star }{\cos \theta _\star }\sin \phi _\star \sin \phi \, \partial _{\theta _\star }h,\\ \partial _{r_\star }\hat{y}& = \sin \theta _\star \cos \phi \,\partial _{r_\star }h,\\ \partial _x\hat{y}& = \sin (h(r_\star ,\theta _\star )) +\frac{\sin \theta _\star }{\cos \theta _\star }\cos \phi _\star \cos \phi \, \partial _{\theta _\star }h,\\ \partial _y\hat{y}& = \cos (h(r_\star ,\theta _\star )) +\frac{\sin \theta _\star }{\cos \theta _\star }\sin \phi _\star \cos \phi \, \partial _{\theta _\star }h. \end{aligned}$$

Note that these are smooth functions and that at \(\theta _\star =0\) they are independent of \(\phi\) and \(\phi _\star\). The differentiability of \((t,r_\star ,\hat{x},\hat{y})\mapsto (t,r_\star ,x,y)\) follows in a similar way.

Regularity of the spheres given by the intersection of hypersurfaces \(u=\hbox {constant}\) and \(v=\hbox {constant}\).

It is obvious that the two atlases \(\{(t,r_\star ,\theta _\star ,\phi _\star ),(t,r_\star ,x,y)\}\) and \(\mathcal{A}_{\tiny \hbox { DN}}=\{(u,v,\theta _\star ,\phi _\star ),(u,v,x,y)\}\) are compatible. So, combining the conclusions of the previous paragraphs, the two atlases \(\mathcal{A}_{\tiny \hbox { BL}}=\{(t,r,\theta ,\phi ),(t,r,\tilde{x},\tilde{y})\}\) and \(\mathcal{A}_{\tiny \hbox { DN}}\) are compatible. The spheres given by the intersection of hypersurfaces \(u=\hbox {constant}\) and \(v=\hbox {constant}\) are smooth in the atlas \(\mathcal{A}_{\tiny \hbox { DN}}\) by definition. Therefore, they are smooth in the atlas \(\mathcal{A}_{\tiny \hbox { BL}}\). This proves

Theorem 2.23

The topological two-spheres given by the intersection of hypersurfaces \(u=\hbox {constant}\) and \(v=\hbox {constant}\) are \(C^\infty\) in the Boyer–Lindquist coordinates.

2.3 Coordinates at the horizons

2.3.1 The decay of \(\Omega ^2\) at the horizons

Recall that the surface gravities of the horizons are given by

$$\begin{aligned} \kappa _-& = -\,\frac{\Lambda }{6}\frac{(r_--r_c)(r_--r_+)(r_--r_n)}{r_-^2+a^2},\nonumber \\ \kappa _+& = -\,\frac{\Lambda }{6}\frac{(r_+-r_c)(r_+-r_-)(r_+-r_n)}{r_+^2+a^2},\nonumber \\ \kappa _c& = -\,\frac{\Lambda }{6}\frac{(r_c-r_+)(r_c-r_-)(r_c-r_n)}{r_c^2+a^2} \end{aligned}$$
(73)

(confirm the formula for \(\kappa _-\) with Example A.3).

Lemma 2.24

Given \(C_R\in {{\mathbb {R}}}\), there exist \(c,C>0\) such that

$$\begin{aligned}{} & {} ce^{2\kappa _-r_\star }\le \Omega ^2\le Ce^{2\kappa _-r_\star }\ \hbox {for}\ r_\star \ge C_R, \end{aligned}$$
(74)
$$\begin{aligned}{} & {} ce^{2\kappa _+r_\star }\le \Omega ^2\le Ce^{2\kappa _+r_\star }\ \hbox {for}\ r_\star \le C_R.\, \end{aligned}$$
(75)

Proof

The formula

$$\begin{aligned} \frac{\partial r}{\partial r_{\star }}=\frac{\Delta _{\theta } \Delta _r \sqrt{(r^2+a^2)^2-a^2 \sin ^2\theta _{\star } \Delta _r}}{\Upsilon } \end{aligned}$$

shows that \(\frac{\partial r}{\partial r_{\star }}(r=r_-,\theta )=0\). To write the linear approximation for this function, we start by calculating

$$\begin{aligned} \partial _r\left( \frac{\partial r}{\partial r_{\star }}\right) \Big |_{r=r_-}& = \frac{\Delta _{\theta }(\partial _r \Delta _r) \sqrt{(r^2+a^2)^2-a^2 \sin ^2\theta _{\star } \Delta _r}}{\Upsilon }\Bigg |_{r=r_-}\\& = \frac{\Delta _{\theta }(\partial _r \Delta _r) (r^2+a^2)}{(r^2+a^2)^2\Delta _{\theta }}\Big |_{r=r_-} =\ \frac{(\partial _r\Delta _r)\Big |_{r=r_-}}{r_-^2+a^2} \ =\ 2\kappa _-. \end{aligned}$$

Hence, we get

$$\begin{aligned} \frac{\partial r}{\partial r_{\star }}=2\kappa _-(r-r_-)+O((r-r_-)^2). \end{aligned}$$

According to Remark 2.1, there exists \(C > 0\) and \((r_\star )_0\) such that for \(r_\star \ge (r_\star )_0\) we have

$$\begin{aligned} ( 1 - C (r - r_-)) (1 + O (r - r_-)) \le 1 \le (1 + C (r - r_-)) (1 + O (r - r_-)). \end{aligned}$$

So, for \(r\ge (r_\star )_0\) we obtain

$$\begin{aligned} \frac{1}{r-r_-}-C\le \frac{1}{\left( r-r_-\right) \left( 1+O \left( r-r_-\right) \right) }\le \frac{1}{r-r_-}+C. \end{aligned}$$

Remembering that \(\kappa _-<0\), it follows that

$$\begin{aligned} \left( \frac{1}{r-r_-}+C\right) \frac{\partial r}{\partial r_{\star }}\le \frac{1}{\left( r-r_-\right) \left( 1+O \left( r-r_-\right) \right) }\frac{\partial r}{\partial r_{\star }}=2 \kappa _-\le \left( \frac{1}{r-r_-}-C\right) \frac{\partial r}{\partial r_{\star }}. \end{aligned}$$

Integrating from \((r_{\star })_0\) to \(r_\star\) yields

$$\begin{aligned}{} & {} \log \left( r(r_{\star },\theta _\star )-r_-\right) -\log \left( r((r_{\star })_0,\theta _\star )-r_-\right) +C \left( r(r_{\star },\theta _\star )-r((r_{\star })_0,\theta _\star )\right) \\{} & {} \qquad \le 2 \kappa _- (r_{\star }-(r_{\star })_0)\le \log \left( r(r_{\star },\theta _\star )-r_-\right) -\log \left( r((r_{\star })_0,\theta _\star )-r_-\right) \\{} & {} \qquad -C \left( r(r_{\star },\theta _\star )-r((r_{\star })_0,\theta _\star )\right) . \end{aligned}$$

These inequalities can be rearranged to

$$\begin{aligned}{} & {} 2 \kappa _- (r_{\star }-(r_{\star })_0) +\log \left( r((r_{\star })_0,\theta _\star )-r_-\right) +C \left( r(r_{\star },\theta _\star )-r((r_{\star })_0,\theta _\star )\right) \\{} & {} \qquad \le \log \left( r(r_{\star },\theta _\star )-r_-\right) \le 2 \kappa _- (r_{\star }-(r_{\star })_0) +\log \left( r((r_{\star })_0,\theta _\star )-r_-\right) \\{} & {} \qquad -C \left( r(r_{\star },\theta _\star )-r((r_{\star })_0,\theta _\star )\right) . \end{aligned}$$

This shows that there exists \(D > 0\) and \((r_\star )_0\) such that

$$\begin{aligned} 2 \kappa _- r_{\star }-D \le \log \left( r(r_{\star },\theta _\star )-r_-\right) \le 2 \kappa _- r_{\star }+D \end{aligned}$$

for \(r_\star \ge (r_\star )_0\). Thus, there exist \(c,C>0\), such that \(r_\star \ge (r_\star )_0\) implies

$$\begin{aligned} ce^{2\kappa _-r_\star }\le r-r_-\le Ce^{2\kappa _-r_\star }. \end{aligned}$$

Let \(C_R\) be a real number. Decreasing c and increasing C if necessary, one sees that

$$\begin{aligned} ce^{2\kappa _-r_\star }\le r-r_-\le Ce^{2\kappa _-r_\star } \end{aligned}$$

for \(C_R\le r_\star \le (r_\star )_0\). Combining the previous two inequalities, they hold for \(r_\star \ge C_R\). Moreover, in this region,

$$\begin{aligned} c(r-r_-)\le -\Delta _r\le C(r-r_-), \end{aligned}$$

for other appropriate constants \(c,C>0\). Hence, given \(C_R\in {{\mathbb {R}}}\), there exist \(c,C>0\) such that

$$\begin{aligned} ce^{2\kappa _-r_\star }\le -\Delta _r\le Ce^{2\kappa _-r_\star } \end{aligned}$$
(76)

for \(r_\star \ge C_R\). As \(\Omega ^2\) is comparable to \(|\Delta _r|\), we conclude that \(\Omega ^2\) is comparable to \(e^{2\kappa _-r_\star }\) in the region \(r_\star \ge C_R\). This proves (74). The proof of (75) is analogous. \(\square\)

Remark 2.25

The function \(\Omega ^2\) given in (19) is obviously smooth on the spheres where u and v are simultaneously constant.

2.3.2 Coordinates at the Cauchy horizon

Let us recall how one may define coordinates to cover the Cauchy horizon. We consider a new smooth coordinate \(v_{{\tiny \mathcal{C}\mathcal{H}^+}}(v)\), with positive derivative, equal to v for \(v\le -1\), satisfying \(v_{{\tiny \mathcal{C}\mathcal{H}^+}}\rightarrow 0\) as \(v\rightarrow +\infty\), and satisfying

$$\begin{aligned} {\textrm{d}}v=e^{-2\kappa _-v}{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}\end{aligned}$$
(77)

for \(v\ge 0\). Moreover, we define

$$\begin{aligned} \phi _{\star ,\mathcal{C}\mathcal{H}^+}=\phi _{\star }-b^{\phi _{\star }}\Big |_{r=r_-}v. \end{aligned}$$

Remark 2.26

\(v_{{\tiny \mathcal{C}\mathcal{H}^+}}\) is also a smooth function of v, and so the change of coordinates \((v,\phi _{\star })\leftrightarrow (v_{{\tiny \mathcal{C}\mathcal{H}^+}},\phi _{\star ,\mathcal{C}\mathcal{H}^+})\) is smooth.

From (20), we see that

$$\begin{aligned} b^{\phi _{\star }}\big |_{r=r_-}=2\frac{\Xi a((r^2+a^2)\Delta _{\theta }-\Delta _r)}{((r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _r)}\Bigg |_{r=r_-}\ =\ \frac{2a\Xi }{r_-^2+a^2}. \end{aligned}$$

For \(v\ge 0\), the differentials of \(\phi _\star\) and \(\phi _{\star ,\mathcal{C}\mathcal{H}^+}\) are related by

$$\begin{aligned} {\textrm{d}}\phi _{\star }={\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}+b^{\phi _{\star }}\Big |_{r=r_-}{\textrm{d}}v ={\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}+b^{\phi _{\star }}\Big |_{r=r_-} e^{-2\kappa _-v}{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}. \end{aligned}$$

For \(v\ge 0\), the expression of the metric (21) in the new coordinates is

$$\begin{aligned} g& = -2\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}\left( {\textrm{d}}u\otimes {\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}+{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}\otimes {\textrm{d}}u\right) \\{} & {} +{g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} {\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\theta _{\star } +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}\theta _{\star }\otimes ({\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}-b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}})\\{} & {} +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}({\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}-b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}})\otimes {\textrm{d}}\theta _{\star }\\{} & {} +{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }} ({\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}-b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}})\otimes ({\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}-b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}), \end{aligned}$$

with

$$\begin{aligned} \Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}& = \Omega ^2e^{-2\kappa _-v},\nonumber \\ b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }& = \left( b^{\phi _{\star }}-b^{\phi _{\star }}\Big |_{r=r_-}\right) e^{-2\kappa _-v}. \end{aligned}$$
(78)

Henceforth we will assume that we are working in the region \(v\ge 0\), our formulas will always refer to this region. To estimate \(b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\) we calculate

$$\begin{aligned} b^{\phi _{\star }}-b^{\phi _{\star }}\Big |_{r=r_-}= -2a\Xi \frac{(r_-^2+a^2\cos ^2\theta )\Delta _r+(r^2+a^2)(r+r_-)(r-r_-)\Delta _{\theta }}{(r_-^2+a^2)((r^2+a^2)^2\Delta _{\theta }-a^2\sin ^2\theta \Delta _r)}. \end{aligned}$$

Using (12) and (76), we estimate

$$\begin{aligned} \left| b^{\phi _{\star }}-b^{\phi _{\star }}\bigg |_{r=r_-}\right| \lesssim e^{2\kappa _-(u+v)}, \end{aligned}$$

for \(u+v\ge C_R\). Thus, inequalities (74) implies the following bounds for \(\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}\) and \(b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\), when \(u+v\ge C_R\):

$$\begin{aligned} e^{2\kappa _-u}\lesssim&\,\,\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}&\lesssim e^{2\kappa _-u},\\&|b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }|&\lesssim e^{2\kappa _-u}. \end{aligned}$$

For a general function f, we have

$$\begin{aligned} f(u,v,\theta _{\star },\phi _\star )=f\left( u,v,\theta _{\star },\phi _{\star ,\mathcal{C}\mathcal{H}^+}+b^{\phi _{\star }}\Big |_{r=r_-}v \right) = \tilde{f} (u,\tilde{v},\theta _{\star },\phi _{\star ,\mathcal{C}\mathcal{H}^+}), \end{aligned}$$

where \(\tilde{f}\) is the function f written in the coordinates \((u,\tilde{v},\theta _{\star },\phi _{\star ,\mathcal{C}\mathcal{H}^+})\) and \({\tilde{v}}=v\). So

$$\begin{aligned} \partial _{\tilde{v}}=\partial _v+b^{\phi _{\star }}\Big |_{r=r_-}\partial _{\phi _\star }. \end{aligned}$$
(79)

We define

$$\begin{aligned} \sigma =e^{-2\kappa _-v}. \end{aligned}$$
(80)

This, (77) and \(\partial _{\phi _{\star }}=\partial _{\phi _{\star ,\mathcal{C}\mathcal{H}^+}}\) imply that

$$\begin{aligned} \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }& = e^{-2\kappa _-v}\left( \partial _{{\tilde{v}}}+\left( b^{\phi _{\star }}-b^{\phi _{\star }}\Big |_{r=r_-}\right) \partial _{\phi _\star }\right) \\& = e^{-2\kappa _-v}\left( \partial _v+b^{\phi _{\star }}\partial _{\phi _\star } \right) \\& = \sigma \left( \partial _v+b^{\phi _{\star }}\partial _{\phi _\star } \right) . \end{aligned}$$

We can write the vector field \(\partial _t\) using the coordinates at the Cauchy horizon as

$$\begin{aligned} \partial _t& = \frac{1}{2}\left( \partial _{\tilde{v}}-b^{\phi _{\star }}\Big |_{r=r_-}\partial _{\phi _\star }\right) -\,\frac{1}{2}\partial _u\\& = \frac{e^{2\kappa _-v}}{2}\left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) -\,\frac{1}{2}b^{\phi _{\star }}\partial _{\phi _\star }-\,\frac{1}{2}\partial _u. \end{aligned}$$

The vector field \(\partial _t\) is not null on the Cauchy horizon. A Killing vector field which is null on the Cauchy horizon is

$$\begin{aligned} Z& = -\partial _t-\,\frac{1}{2}b^{\phi _{\star }}|_{r=r_-}\partial _{\phi _\star }\nonumber \\& = \frac{1}{2}\partial _u+\frac{1}{2}\left( b^{\phi _{\star }}-b^{\phi _{\star }}|_{r=r_-}\right) \partial _{\phi _\star }-\,\frac{e^{2\kappa _-v}}{2}(\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star })\nonumber \\& = \frac{1}{2}\partial _u+\frac{1}{2}\left( b^{\phi _{\star }}-b^{\phi _{\star }}|_{r=r_-}\right) \partial _{\phi _\star }-\,\frac{\Omega ^2}{2}\frac{\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}. \end{aligned}$$
(81)

2.3.3 Coordinates at the event horizon

We consider a new smooth coordinate \(u_{\mathcal{H}^+}(u)\), with positive derivative, equal to u for \(u\ge 1\), satisfying \(u_{\mathcal{H}^+}\rightarrow 0\) as \(u\rightarrow -\infty\), and satisfying

$$\begin{aligned} {\textrm{d}}u=e^{-2\kappa _+u}{\,\textrm{d}}u_{\mathcal{H}^+}\end{aligned}$$

for \(u\le 0\). Moreover, we define \(v_{\mathcal{H}^+}=v\) and

$$\begin{aligned} \phi _{\star ,\mathcal{H}^+}=\phi _\star -b^{\phi _{\star }}|_{r=r_+}v. \end{aligned}$$

Remark 2.27

\(u_{\mathcal{H}^+}\) is also a smooth function of u, and the change of coordinates \((v,\phi _{\star })\leftrightarrow (v_{\mathcal{H}^+},\phi _{\star ,\mathcal{H}^+})\) is smooth.

Note that

$$\begin{aligned} b^{\phi _{\star }}|_{r=r_+}=\frac{2a\Xi }{r_+^2+a^2}. \end{aligned}$$

For \(u\le 0\), we may write the metric as

$$\begin{aligned} g& = -2\Omega ^2_{{\tiny \mathcal{H}^+}}\left( {\textrm{d}}u_{\mathcal{H}^+}\otimes {\textrm{d}}v_{\mathcal{H}^+}+{\,\textrm{d}}v_{\mathcal{H}^+}\otimes {\textrm{d}}u_{\mathcal{H}^+}\right) \\{} & {} +{g \hspace{-4.83322pt}/}_{\theta _{\star }\theta _{\star }} {\,\textrm{d}}\theta _{\star }\otimes {\textrm{d}}\theta _{\star } +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}{\,\textrm{d}}\theta _{\star }\otimes ({\textrm{d}}\phi _{\star ,\mathcal{H}^+}-b_{{\tiny \mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{\mathcal{H}^+}) \\{} & {} +{g \hspace{-4.83322pt}/}_{\theta _{\star }\phi _{\star }}({\textrm{d}}\phi _{\star ,\mathcal{H}^+}-b_{{\tiny \mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{\mathcal{H}^+})\otimes {\textrm{d}}\theta _{\star }\\{} & {} +{g \hspace{-4.83322pt}/}_{\phi _{\star }\phi _{\star }} ({\textrm{d}}\phi _{\star ,\mathcal{H}^+}-b_{{\tiny \mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{\mathcal{H}^+})\otimes ({\textrm{d}}\phi _{\star ,\mathcal{H}^+}-b_{{\tiny \mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{\mathcal{H}^+}), \end{aligned}$$

with

$$\begin{aligned} \Omega ^2_{{\tiny \mathcal{H}^+}}& = \Omega ^2e^{-2\kappa _+u},\\ b_{{\tiny \mathcal{H}^+}}^{\phi _\star }& = b^{\phi _{\star }}-b^{\phi _{\star }}|_{r=r_+}. \end{aligned}$$

Henceforth we will assume that we are working in the region \(u\le 0\), our formulas will always refer to this region. For a general function f, we have

$$\begin{aligned} f(u,v,\theta _{\star },\phi _\star )=f\left( u(u_{\mathcal{H}^+}),v,\theta _{\star },\phi _{\star ,\mathcal{H}^+}+b^{\phi _{\star }}\big |_{r=r_+}v \right) = \tilde{f} (u_{\mathcal{H}^+},v_{\mathcal{H}^+},\theta _{\star },\phi _{\star ,\mathcal{H}^+}), \end{aligned}$$

where \(\tilde{f}\) is the function f written in the coordinates \((u_{\mathcal{H}^+},v_{\mathcal{H}^+},\theta _{\star },\phi _{\star ,\mathcal{H}^+})\). So, defining

$$\begin{aligned} \varsigma =e^{-2\kappa _+u}, \end{aligned}$$
(82)

we get

$$\begin{aligned} \partial _{u_{\mathcal{H}^+}}& = e^{-2\kappa _+u}\partial _u\ =\ \varsigma \partial _u,\\ \partial _{v_{\mathcal{H}^+}}& = \partial _v+b^{\phi _{\star }}\big |_{r=r_+}\partial _{\phi _\star }. \end{aligned}$$

This, (77) and \(\partial _{\phi _{\star }}=\partial _{\phi _{\star ,\mathcal{H}^+}}\) imply that

$$\begin{aligned} \partial _v+b^{\phi _{\star }}\partial _{\phi _\star }& = \partial _{v_{\mathcal{H}^+}}+b_{{\tiny \mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }. \end{aligned}$$

A Killing vector field which is null on the event horizon is

$$\begin{aligned} W=\partial _t+\frac{1}{2}b^{\phi _{\star }}\Big |_{r=r_+}\partial _{\phi _\star }= \frac{1}{2}\partial _{v_{\mathcal{H}^+}}-\,\frac{e^{2\kappa _+u}}{2}\partial _{u_{\mathcal{H}^+}}. \end{aligned}$$
(83)

The value

$$\begin{aligned} \frac{1}{2}b^{\phi _{\star }}\Big |_{r=r_+} =\frac{a\Xi }{r_+^2+a^2} \end{aligned}$$
(84)

is the angular velocity \(\Omega _H\) on the event horizon.

3 The energy of the solutions of the wave equation

We will use the vector field method to study the energy of solutions of the wave equation which have compact support on \(\mathcal{H}^+\). As is well known, the method, used by Morawetz [26], John [20], Klainerman [21, 22], Dafermos [8,9,10,11,12] and Rodnianski [11, 22], among many others, consists in applying the Divergence Theorem to some currents obtained by contracting the energy-momentum tensor \(T_{\mu \nu }\) with appropriate vector field multipliers constructed specifically according to each region of spacetime.

We refer to the region close to the Cauchy horizon as the blue-shift region, and the region close to the event horizon as the red-shift region. We call the intermediate region the no-shift region. A very general construction of red-shift vector fields on general spacetimes which contain Killing horizons with positive surface gravity is carried out in the lecture notes [11]. Here we perform the computations explicitly in double null coordinates.

The blue-shift vector field (85) is constructed using the vector field Y in Lemma 3.1, and the red-shift vector field (96) is constructed using the vector field V in Lemma 3.2. We go on to calculate the covariant derivative of Y, \(\nabla ^\mu Y^\nu\), and the scalar current associated to Y, \(T_{\mu \nu }\nabla ^\mu Y^\nu\). We obtain the usual inequalities for the currents associated to the blue-shift vector field, and for the currents associated to the red-shift vector field. We finish with Theorem 3.5, which is Sbierski’s result, for the Reissner–Nordström and Kerr spacetimes, applied to Kerr–Newman–de Sitter spacetimes.

3.1 The blue-shift and red-shift vector fields

3.1.1 Construction

Here the blue-shift vector field is defined to be

$$\begin{aligned} N_b=Y+Z, \end{aligned}$$
(85)

where Y is given in

Lemma 3.1

Let \(\iota \in {{\mathbb {R}}}^+\) be given. The initial value problem

$$\begin{aligned} \nabla _YY& = -\iota (Y+Z), \end{aligned}$$
(86)
$$\begin{aligned} Y|_{\mathcal{C}\mathcal{H}^+}& = \frac{\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\end{aligned}$$
(87)

(Z as in (81)) has a unique time invariant solution, Y, defined in a neighborhood of the Cauchy horizon, i.e. defined for \(r_\star\) sufficiently large.

Proof

(a):

(Y time invariant.) The vector field \({\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right)\) commutes with \(\partial _t\). In fact, we have

$$\begin{aligned} \left[ \partial _t,\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\right] = \left[ \partial _t,\frac{1}{\Omega ^2}\partial _{\tilde{v}}\right] =\frac{1}{\Omega ^2}\left[ \frac{1}{2}\partial _v-\,\frac{1}{2}\partial _u, \left( \partial _v+b^{\phi _{\star }}|_{r=r_-}\partial _{\phi _\star }\right) \right] =0 \end{aligned}$$

and

$$\begin{aligned} \frac{b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}=\frac{b^{\phi _{\star }}-b^{\phi _{\star }}|_{r=r_-}}{\Omega ^2} \ \Rightarrow \ \left[ \partial _t,\frac{b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{\phi _\star }\right] =0. \end{aligned}$$

So, any Y of the form

$$\begin{aligned} Y& = \tilde{f}\partial _u+\tilde{g}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }, \end{aligned}$$

with

$$\begin{aligned} \tilde{f}=\tilde{f}(r,\theta ),\ \tilde{g}=\tilde{g}(r,\theta ),\ \tilde{h}=\tilde{h}(r,\theta ),\ \tilde{{\jmath }}=\tilde{{\jmath }}(r,\theta ), \end{aligned}$$

commutes with \(\partial _t\).

(b):

(The differential equation.) Expanding the left-hand side of (86), we get

$$\begin{aligned} \nabla _YY& = \tilde{f}\partial _{r_\star }\tilde{f}\partial _u+\tilde{f}\partial _{r_\star }\tilde{g}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{f}\partial _{r_\star }\tilde{h}\partial _{\theta _\star }+\tilde{f}\partial _{r_\star }\tilde{{\jmath }}\partial _{\phi _\star }\\{} & {} +\tilde{f}^2\nabla _{\partial _u}\partial _u+\tilde{f}\tilde{g}\nabla _{\partial _u}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{f}\tilde{h}\nabla _{\partial _u}\partial _{\theta _\star }+\tilde{f}\tilde{{\jmath }}\nabla _{\partial _u}\partial _{\phi _\star }\\{} & {} +\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{f}\partial _u+\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{g}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{h}\partial _{\theta _\star }\\{} & {} +\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{{\jmath }}\partial _{\phi _\star }\\{} & {} +\tilde{f}\tilde{g}\nabla _{{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) }\partial _u+\tilde{g}\tilde{h}\nabla _{{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) }\partial _{\theta _\star }\\{} & {} +\tilde{g}\tilde{{\jmath }}\nabla _{{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) }\partial _{\phi _\star }\\{} & {} +\tilde{h}\partial _{\theta _\star }\tilde{f}\partial _u+\tilde{h}\partial _{\theta _\star }\tilde{g}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{h}\partial _{\theta _\star }\tilde{h}\partial _{\theta _\star }+\tilde{h}\partial _{\theta _\star }\tilde{{\jmath }}\partial _{\phi _\star }\\{} & {} +\tilde{f}\tilde{h}\nabla _{\partial _{\theta _\star }}\partial _u+\tilde{g}\tilde{h}\nabla _{\partial _{\theta _\star }}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{h}^2\nabla _{\partial _{\theta _\star }}\partial _{\theta _\star }+\tilde{h}\tilde{{\jmath }}\nabla _{\partial _{\theta _\star }}\partial _{\phi _\star }\\{} & {} +\tilde{f}\tilde{{\jmath }}\nabla _{\partial _{\phi _\star }}\partial _u+\tilde{g}\tilde{{\jmath }}\nabla _{\partial _{\phi _\star }}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{h}\tilde{{\jmath }}\nabla _{\partial _{\phi _\star }}\partial _{\theta _\star }+\tilde{{\jmath }}^2\nabla _{\partial _{\phi _\star }}\partial _{\phi _\star }. \end{aligned}$$

We used (109) to eliminate the term in \(\tilde{g}^2\). This initial value problem (86), (87) is equivalent to a system of four equations, for the \(\partial _u\), \({\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right)\), \(\partial _{\theta _\star }\) and \(\partial _{\phi _\star }\) components of each side, for the four unknowns \(\tilde{f}\), \(\tilde{g}\), \(\tilde{h}\) and \(\tilde{{\jmath }}\), with

$$\begin{aligned} \tilde{f}(r_-,\theta )=0,\quad \tilde{g}(r_-,\theta )=1,\quad \tilde{h}(r_-,\theta )=0,\quad \tilde{{\jmath }}(r_-,\theta )=0. \end{aligned}$$
(88)

The system reads

$$\begin{aligned} \tilde{f}\partial _{r_\star }\tilde{f}+\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{f}+\tilde{h}\partial _{\theta _\star }\tilde{f}& = \ldots ,\\ \tilde{f}\partial _{r_\star }\tilde{g}+\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{g}+\tilde{h}\partial _{\theta _\star }\tilde{g}& = \ldots ,\\ \tilde{f}\partial _{r_\star }\tilde{h}+\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{h}+\tilde{h}\partial _{\theta _\star }\tilde{h}& = \ldots ,\\ \tilde{f}\partial _{r_\star }\tilde{{\jmath }}+\tilde{g}\frac{\sigma }{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{r_\star }\tilde{{\jmath }}+\tilde{h}\partial _{\theta _\star }\tilde{{\jmath }}& = \ldots , \end{aligned}$$

where the right-hand sides involve the Christoffel symbols of the metric, \(\partial _{r_\star }b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\), \(\partial _{\theta _\star }b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\), and \(\tilde{f}\), \(\tilde{g}\), \(\tilde{h}\) and \(\tilde{{\jmath }}\), but do not involve any derivatives of these last four functions. The system may be written as

$$\begin{aligned} \mathcal{V}\cdot \tilde{f}& = \ldots ,\end{aligned}$$
(89)
$$\begin{aligned} \mathcal{V}\cdot \tilde{g}& = \ldots ,\end{aligned}$$
(90)
$$\begin{aligned} \mathcal{V}\cdot \tilde{h}& = \ldots ,\end{aligned}$$
(91)
$$\begin{aligned} \mathcal{V}\cdot \,\tilde{{\jmath }}& = \ldots , \end{aligned}$$
(92)

where

$$\begin{aligned}{} & {} \mathcal{V}= \left( \left( \frac{1}{\Delta _r}\frac{\partial r}{\partial r_\star }\right) \left( \tilde{f}\Delta _r+\tilde{g}\left( -\,\frac{\Upsilon }{\rho ^2\Delta _\theta }\right) \right) +\tilde{h}\frac{\partial r}{\partial \theta _\star }\right) \frac{\partial }{\partial r}+\\{} & {} \quad \left( \left( \frac{1}{\Delta _r}\frac{\partial \theta }{\partial r_\star }\right) \left( \tilde{f}\Delta _r+\tilde{g}\left( -\,\frac{\Upsilon }{\rho ^2\Delta _\theta }\right) \right) +\tilde{h}\frac{\partial \theta }{\partial \theta _\star }\right) \frac{\partial }{\partial \theta }. \end{aligned}$$

We used

$$\begin{aligned} \frac{\sigma \Delta _r}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}=-\,\frac{\Upsilon }{\rho ^2\Delta _\theta }, \end{aligned}$$

which we obtain from (19), (78) and (80).

(c):

(An auxiliary calculation.) In the next step we will use the following identity. We mention that it implies that \(\tilde{h}\) is not identically equal to zero. Using (111) and (112), we have

$$\begin{aligned}{} & {} {\,\textrm{d}}\theta _\star \left( \nabla _{\partial _u}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\nabla _{{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) }\partial _u\right) =2{g \hspace{-4.83322pt}/}^{\theta _\star \theta _\star }\frac{\partial _{\theta _\star }(\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}})}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}. \end{aligned}$$
(93)
(d):

(The characteristics do not cross the boundary.) We now check that

$$\begin{aligned} \tilde{h}(r,0)=0\qquad \hbox {and}\qquad \tilde{h}\left( r,\frac{\pi }{2}\right) =0. \end{aligned}$$
(94)

This implies

$$\begin{aligned} \mathcal{V}^\theta (r,0)=0\qquad \hbox {and}\qquad \mathcal{V}^\theta \left( r,\frac{\pi }{2}\right) =0, \end{aligned}$$
(95)

and guarantees that the characteristics of our differential equations do not leave the region, \([r_-,r_+]\times \left[ 0,\frac{\pi }{2}\right]\), where we want to solve our system. It also guarantees that Y is well defined when \(\theta =0\), notwithstanding \(\partial _{\theta _\star }\) not being well defined when \(\theta =0\). The vanishing of \(\tilde{h}\) at \(\theta =\frac{\pi }{2}\) can also be seen as a consequence of the symmetry of our problem under the reflection \(\theta \mapsto \pi -\theta\), which implies that Y should not have any component in the \(\partial _{\theta _\star }\) direction at the equators of the spheres where u and v are both constant. The right hand side of (91) consists of a sum of terms which we divide into two parts. The first part consists of sum of the eight summands that have \(\tilde{h}\) as a factor. The term

$$\begin{aligned} {\,\textrm{d}}\theta _\star (-\iota (Y+Z))=-\iota {\,\textrm{d}}\theta _\star (Y) =-\iota \tilde{h} \end{aligned}$$

is proportional to \(\tilde{h}\). The second part consists of the sum of the remaining eight summands, which are \({\,\textrm{d}}\theta _\star\) applied to

$$\begin{aligned}{} & {} -\tilde{f}^2\nabla _{\partial _u}\partial _u-\tilde{f}\tilde{g}\nabla _{\partial _u}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) -2\tilde{f}\tilde{{\jmath }}\nabla _{\partial _u}\partial _{\phi _\star }\\{} & {} -\tilde{f}\tilde{g}\nabla _{{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) }\partial _u-2\tilde{g}\tilde{{\jmath }}\nabla _{{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) }\partial _{\phi _\star }-\tilde{{\jmath }}^2\nabla _{\partial _{\phi _\star }}\partial _{\phi _\star }. \end{aligned}$$

Taking into account (93), the second part is

$$\begin{aligned}{} & {} -\tilde{f}^2\Gamma _{uu}^{\theta _\star }-4\tilde{f}\tilde{g}\frac{{g \hspace{-4.83322pt}/}^{\theta _\star \theta _\star }}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}\partial _{\theta _\star }(\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}})-2\tilde{f}\tilde{{\jmath }}\Gamma _{u\phi _{\star ,\mathcal{C}\mathcal{H}^+}}^{\theta _\star }\\{} & {} -2\tilde{g}\tilde{{\jmath }}\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}} \left( \Gamma _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}}^{\theta _\star }+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\Gamma _{\phi _{\star ,\mathcal{C}\mathcal{H}^+}\phi _{\star ,\mathcal{C}\mathcal{H}^+}}^{\theta _\star }\right) -\tilde{{\jmath }}^2\Gamma _{\phi _{\star ,\mathcal{C}\mathcal{H}^+}\phi _{\star ,\mathcal{C}\mathcal{H}^+}}^{\theta _\star }. \end{aligned}$$

At \(\theta _\star =0\) and at \(\theta _\star =\frac{\pi }{2}\) this sum is zero because each of the terms is equal to zero. Indeed, all terms are a product of a differentiable function by \(\sin (2\theta _\star )\). This is easy to check. Let us exemplify this assertion with one of the less immediate terms to analyze, the one which contains

$$\begin{aligned} \Gamma _{u\phi _{\star ,\mathcal{C}\mathcal{H}^+}}^{\theta _\star }= \frac{{g \hspace{-4.83322pt}/}^{\theta _\star \theta _\star }}{2} \partial _{r_\star }{g \hspace{-4.83322pt}/}_{\theta _\star \phi _\star } +\frac{{g \hspace{-4.83322pt}/}^{\theta _\star \phi _\star }}{2} \partial _{r_\star }{g \hspace{-4.83322pt}/}_{\phi _\star \phi _\star } \end{aligned}$$

(see (106)). Since both \({g \hspace{-4.83322pt}/}_{\theta _\star \phi _\star }\) and \({g \hspace{-4.83322pt}/}^{\theta _\star \phi _\star }\) contain \(\sin (2\theta _\star )\), so does this Christoffel symbol. Let us examine in more detail (91) for points (r, 0) and \(\left( r,\frac{\pi }{2}\right)\). Since \(\partial _{r_\star }\theta\) also contains a factor \(\sin (2\theta _\star )\), over these two segments, \([r_-,r_+)\times \{0\}\) and \([r_-,r_+)\times \{\frac{\pi }{2}\}\), (91) reads

$$\begin{aligned} \mathcal{V}^r\partial _r\tilde{h}=-\tilde{h}\frac{\partial \theta }{\partial \theta _\star }\frac{\partial \tilde{h}}{\partial \theta }+\tilde{h}\times \hbox {smooth function}=\tilde{h}\times \hbox {smooth function}. \end{aligned}$$

As

$$\begin{aligned} \tilde{h}(r_-,0)=0\qquad \hbox {and}\qquad \tilde{h}\left( r_-,\frac{\pi }{2}\right) =0 \end{aligned}$$

and \(\mathcal{V}^r\) is not zero (at least initially at \(r_-\), see below), we conclude that, if there exists a solution to our initial value problem, then it must satisfy (94). Using the fact that \(\partial _{r_\star }\theta\) contains a factor \(\sin (2\theta _\star )\) once again, we obtain (95).

(e):

(Existence and uniqueness of solution.) Using (13) and (39), we have

$$\begin{aligned} \mathcal{V}^r(r_-,\theta )=\left( \frac{1}{\Delta _r}\frac{\partial r}{\partial r_\star }\right) \left( -\,\frac{\Upsilon }{\rho ^2\Delta _\theta }\right) =\frac{1}{(r_-^2+a^2)} \left( -\,\frac{(r_-^2+a^2)^2}{\rho ^2}\right) =-\,\frac{(r_-^2+a^2)}{r_-^2+a^2\cos ^2\theta }. \end{aligned}$$

This shows that the segment \(\{r_-\}\times \left[ 0,\frac{\pi }{2}\right]\) is noncharacteristic for our system of four first-order quasilinear partial differential equations. We observe that when the Christoffel symbols \(\Gamma _{v\theta _\star }^{\phi _\star }\) and \(\Gamma _{\theta _\star \phi _\star }^{\phi _\star }\), which blow up at \(\theta _\star =0\) like \(\frac{1}{\sin \theta _\star }\) (see Corollary A.2), appear in the system above, then they appear multiplied by \(\tilde{h}\), which has to vanish to first order at (r, 0). So that the summands where these Christoffel appear are continuous functions. By a standard existence and uniqueness theorem for non characteristic first order quasilinear partial differential equations, we know that our initial value problem has a solution for \((r,\theta )\in [r_-,r_-+\delta ]\times [0,\pi ]\), for some positive \(\delta\). Recall that constructing the solution involves solving the system of ordinary differential equations

$$\begin{aligned} \left\{ \begin{array}{rcl} \dot{r}&{}=&{}\mathcal{V}^r\\ \dot{\theta }&{}=&{}\mathcal{V}^{\theta }\\ \dot{\tilde{f}}&{}=&{}\hbox {right-hand side of}\ (89)\\ \dot{\tilde{g}}&{}=&{}\hbox {right-hand side of}\ (90)\\ \dot{\tilde{h}}&{}=&{}\hbox {right-hand side of}\ (91)\\ \dot{\tilde{{\jmath }}}&{}=&{}\hbox {right-hand side of}\ (92) \end{array} \right. \qquad \hbox {with}\qquad \left\{ \begin{array}{rcl} r(0)&{}=&{}r_-\\ \theta (0)&{}=&{}\theta _0\\ \tilde{f}(0)&{}=&{}0\\ \tilde{g}(0)&{}=&{}1\\ h(0)&{}=&{}0\\ \tilde{{\jmath }}(0)&{}=&{}0 \end{array} \right. . \end{aligned}$$

We know from Remark 2.1 that, for \(r_\star\) sufficiently large, r is close to \(r_-\). So we have the existence of a solution for large \(r_\star\).

\(\square\)

Here the red-shift vector field is defined to be

$$\begin{aligned} N_r=V+W, \end{aligned}$$
(96)

where V is given in

Lemma 3.2

Let \(\iota \in {{\mathbb {R}}}^+\) be given. The initial value problem

$$\begin{aligned} \nabla _VV={} & {} -\iota (V+W),\\ V|_{\mathcal{H}^+}& = \frac{\partial _{u_{\mathcal{H}^+}}}{\Omega ^2_{{\tiny \mathcal{H}^+}}} \end{aligned}$$

(W as in (83)) has a unique time invariant solution, V, defined in a neighborhood of the event horizon, i.e. defined for \(r_\star\) sufficiently negative (i.e. for \(r_\star <-C\) with C sufficiently large).

Proof

Choose V of the form

$$\begin{aligned} V& = \tilde{f}{\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{H}^+}}}}\partial _{u_{\mathcal{H}^+}}+\tilde{g} \left( \partial _{v_{\mathcal{H}^+}}\!\!+b_{{\tiny \mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) +\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }, \end{aligned}$$

with

$$\begin{aligned} \tilde{f}=\tilde{f}(r,\theta ),\ \tilde{g}=\tilde{g}(r,\theta ),\ \tilde{h}=\tilde{h}(r,\theta ),\ \tilde{{\jmath }}=\tilde{{\jmath }}(r,\theta ),\\ \tilde{f}(r_+,\theta )=1,\ \tilde{g}(r_+,\theta )=0,\ \tilde{h}(r_+,\theta )=0,\ \tilde{{\jmath }}(r_+,\theta )=0. \end{aligned}$$

\(\square\)

3.1.2 Covariant derivative

One could consider working in the frame \((Z,Y,\partial _{\theta _\star },\partial _{\phi _\star })\) but this is not a good choice because the energy-momentum tensor does not have a simple expression in this frame. So, instead, we define

$$\begin{aligned} \overline{T}=\frac{1}{2}\partial _u\qquad \hbox {and}\qquad \overline{Y}={\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\!\!+b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) = \frac{\partial _v+b^{\phi _{\star }}\partial _{\phi _\star }}{\Omega ^2}, \end{aligned}$$

and work in the frame

$$\begin{aligned} (X_{{\tiny \overline{\texttt {T}}}},X_{{\tiny \overline{\texttt {Y}}}},X_{\theta _\star },X_{\phi _\star }):= (\overline{T},\overline{Y},\partial _{\theta _\star },\partial _{\phi _\star }). \end{aligned}$$

We see that

$$\begin{aligned} Y=2\tilde{f}\overline{T}+\tilde{g}\overline{Y}+\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }. \end{aligned}$$

The dual frame is

$$\begin{aligned} (\omega _{{\tiny \overline{\texttt {T}}}},\omega _{{\tiny \overline{\texttt {Y}}}},\omega _{\theta _\star },\omega _{\phi _\star })& = \left( 2{\,\textrm{d}}u,\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}},{\,\textrm{d}}\theta _\star ,{\,\textrm{d}}\phi _{\star ,\mathcal{C}\mathcal{H}^+}-b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}\right) \\& = \left( 2{\,\textrm{d}}u,\Omega ^2{\,\textrm{d}}v,{\,\textrm{d}}\theta _\star ,{\,\textrm{d}}\phi _\star -b^{\phi _{\star }}{\,\textrm{d}}v \right) . \end{aligned}$$

Calculation of the covariant derivative of Y.

The covariant derivative of Y is

$$\begin{aligned} \nabla Y& = \omega _{{\tiny \overline{\texttt {T}}}}^\sharp \otimes \nabla _{\overline{T}}Y +\omega _{{\tiny \overline{\texttt {Y}}}}^\sharp \otimes \nabla _{\overline{Y}}Y +\omega _{\theta _\star }^\sharp \otimes \nabla _{\partial _{\theta _\star }}Y +\omega _{\phi _\star }^\sharp \otimes \nabla _{\partial _{\phi _\star }}Y. \end{aligned}$$

One readily checks that the metric dual basis to \((\omega _{{\tiny \overline{\texttt {T}}}},\omega _{{\tiny \overline{\texttt {Y}}}},\omega _{\theta _\star },\omega _{\phi _\star })\) is

$$\begin{aligned} (\omega _{{\tiny \overline{\texttt {T}}}}^\sharp ,\omega _{{\tiny \overline{\texttt {Y}}}}^\sharp , \omega _{\theta _\star }^\sharp ,\omega _{\phi _\star }^\sharp )& = (-\overline{Y},-\overline{T}, {g \hspace{-4.83322pt}/}^{\theta _\star \theta _\star }\partial _{\theta _\star } +{g \hspace{-4.83322pt}/}^{\theta _\star \phi _\star }\partial _{\phi _\star }, {g \hspace{-4.83322pt}/}^{\theta _\star \phi _\star }\partial _{\theta _\star } +{g \hspace{-4.83322pt}/}^{\phi _\star \phi _\star }\partial _{\phi _\star })\\& = (-\overline{Y},-\overline{T}, \partial ^{\theta _\star }, \partial ^{\phi _\star }). \end{aligned}$$

Using the fact that \(2\overline{T}=\partial _u=\partial _{r_\star }= (\partial _{r_\star }r)\partial _r+(\partial _{r_\star }\theta )\partial _\theta\), (88) and (112), we obtain

$$\begin{aligned} \nabla _{\overline{T}}Y& = \nabla _{\overline{T}}\left( 2\tilde{f}\overline{T}+\tilde{g}\overline{Y}+\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }\right) \nonumber \\& = \nabla _{\overline{T}}\overline{Y}+\overline{O}(r-r_-)\nonumber \\& = -\kappa _-\overline{Y}+a^{\theta _\star }\partial _{\theta _\star }+ a^{\phi _\star }\partial _{\phi _\star } +\tilde{O}(r-r_-), \end{aligned}$$
(97)

where

$$\begin{aligned} \tilde{O}(r-r_-)=O(r-r_-)\overline{T}+O(r-r_-)\overline{Y}+O(r-r_-)\partial _{\theta _\star } +O(r-r_-)\partial _{\phi _\star }. \end{aligned}$$

The values of \(a^{\theta _\star }\) and \(a^{\phi _\star }\) can be read off from (112). Using (86) and the formulas in "Appendix  A", we get

$$\begin{aligned} \nabla _{\overline{Y}}Y& = \nabla _YY-\nabla _{Y-\overline{Y}}Y\nonumber \\& = \nabla _YY-\nabla _{\left( 2\tilde{f}\overline{T}+(\tilde{g}-1)\overline{Y}+\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }\right) } \left( 2\tilde{f}\overline{T}+\tilde{g}\overline{Y}+\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }\right) \nonumber \\& = -\iota (Y+Z)+\overline{O}(r-r_-)\nonumber \\& = -\iota (\overline{Y}+\overline{T})+\overline{O}(r-r_-), \end{aligned}$$
(98)

where

$$\begin{aligned} \overline{O}(r-r_-)=O(r-r_-)\overline{T}+O(r-r_-)\overline{Y}+O(r-r_-)\partial _{\theta _\star } +O(r-r_-)\frac{1}{\sin \theta }\partial _{\phi _\star }. \end{aligned}$$

The factor \(\frac{1}{\sin \theta }\) in front of \(\partial _{\phi _\star }\) in the last summand arises from \(\Gamma _{v\theta _\star }^{\phi _\star }\) and \(\Gamma _{\theta _\star \phi _\star }^{\phi _\star }\). Using the fact that \(\partial _{\theta _\star }= (\partial _{\theta _\star }r)\partial _r +(\partial _{\theta _\star }\theta )\partial _\theta\) and (113), we obtain

$$\begin{aligned} \nabla _{\partial _{\theta _\star }}Y& = \nabla _{\partial _{\theta _\star }}\left( 2\tilde{f}\overline{T}+\tilde{g}\overline{Y}+\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }\right) \nonumber \\& = \nabla _{\partial _{\theta _\star }}\overline{Y}+\overline{O}(r-r_-)\nonumber \\& = -a_{\theta _\star }\overline{Y} +h^{\ \ \theta _\star }_{\theta _\star }\partial _{\theta _\star }+ h^{\ \ \phi _\star }_{\theta _\star }\partial _{\phi _\star } +\overline{O}(r-r_-). \end{aligned}$$
(99)

The values of \(h^{\ \ \theta _\star }_{\theta _\star }\) and \(h^{\ \ \phi _\star }_{\theta _\star }\) can be read off from (113). Finally, we have

$$\begin{aligned} \nabla _{\partial _{\phi _\star }}Y& = \nabla _{\partial _{\phi _\star }}\left( 2\tilde{f}\overline{T}+\tilde{g}\overline{Y}+\tilde{h}\partial _{\theta _\star }+\tilde{{\jmath }}\partial _{\phi _\star }\right) \nonumber \\& = \nabla _{\partial _{\phi _\star }}\overline{Y}+\overline{O}(r-r_-)\nonumber \\& = -a_{\phi _\star }\overline{Y} +h^{\ \ \theta _\star }_{\phi _\star }\partial _{\theta _\star }+ h^{\ \ \phi _\star }_{\phi _\star }\partial _{\phi _\star } +\hat{O}(r-r_-). \end{aligned}$$
(100)

The values of \(h^{\ \ \theta _\star }_{\phi _\star }\) and \(h^{\ \ \phi _\star }_{\phi _\star }\) can be read off from (114). Here

$$\begin{aligned} \hat{O}(r-r_-)=O(r-r_-)\sin \theta \overline{T}+O(r-r_-)\sin \theta \overline{Y}+O(r-r_-) \sin \theta \partial _{\theta _\star } +O(r-r_-)\partial _{\phi _\star }. \end{aligned}$$

Note that although \(\Gamma _{\theta _\star \phi _\star }^{\phi _\star }\) does contain the factor \(\frac{1}{\sin \theta }\), in the last calculation this Christoffel symbol appears multiplied by \(\tilde{h}\) which vanishes to first order at \(\theta _\star =0\) and \(\theta _\star =\pi\). So, the last equality is a consequence of Lemma A.1 and of the fact that \(\tilde{f}\), \(\tilde{g}\), \(\tilde{h}\) and \(\tilde{{\jmath }}\) do not depend on \(\phi _\star\). Indeed, the components of \(\nabla _{\partial _{\phi _\star }}X_\dagger\) in \(\overline{T}\), \(\overline{Y}\) and \(\partial _{\theta _\star }\) all contain the factor \(\sin \theta\), for \(\dagger \,\in \,\{{ \overline{\texttt {T}}},{ \overline{\texttt {Y}}},\theta _\star ,\phi _\star \}\). The expressions (97)–(100) above correspond to [11, (19)–(22)]. Combining the previous results, we can write the covariant derivative of Y as

$$\begin{aligned} \nabla Y& = -\overline{Y}\otimes \left( -\kappa _-\overline{Y}+a^{\theta _\star }\partial _{\theta _\star }+ a^{\phi _\star }\partial _{\phi _\star } \right) +\overline{Y}\otimes \tilde{O}(r-r_-) \\{} & {} -\overline{T}\otimes \left( -\iota (\overline{Y}+\overline{T}) \right) +\overline{T}\otimes \overline{O}(r-r_-)\\{} & {} +\partial ^{\theta _\star }\otimes \left( -a_{\theta _\star }\overline{Y} +h_{\theta _\star \theta _\star }\partial ^{\theta _\star }+ h_{\theta _\star \phi _\star }\partial ^{\phi _\star } \right) +\partial ^{\theta _\star }\otimes \overline{O}(r-r_-)\\{} & {} +\partial ^{\phi _\star }\otimes \left( -a_{\phi _\star }\overline{Y} +h_{\phi _\star \theta _\star }\partial ^{\theta _\star }+ h_{\phi _\star \phi _\star }\partial ^{\phi _\star } \right) +\partial ^{\phi _\star }\otimes \hat{O}(r-r_-)\\& = \kappa _-\overline{Y}\otimes \overline{Y} -\overline{Y}\otimes (a^{\theta _\star }\partial _{\theta _\star }) -\overline{Y}\otimes (a^{\phi _\star }\partial _{\phi _\star })\\{} & {} +\iota \overline{T}\otimes \overline{Y}+\iota \overline{T}\otimes \overline{T}\\{} & {} -(a^{\theta _\star }\partial _{\theta _\star })\otimes \overline{Y} -(a^{\phi _\star }\partial _{\phi _\star })\otimes \overline{Y}\\{} & {} +h^{\theta _\star \theta _\star }\partial _{\theta _\star }\otimes \partial _{\theta _\star } +h^{\theta _\star \phi _\star }\partial _{\theta _\star }\otimes \partial _{\phi _\star } +h^{\phi _\star \theta _\star }\partial _{\phi _\star }\otimes \partial _{\theta _\star } +h^{\phi _\star \phi _\star }\partial _{\phi _\star }\otimes \partial _{\phi _\star }\\{} & {} +\overline{\overline{O}}(r-r_-). \end{aligned}$$

We may write the error term as

$$\begin{aligned} \overline{\overline{O}}(r-r_-)& = \sum _{\dagger ,\ddagger \, \in \,\{{\tiny \overline{\texttt {T}}},{\tiny \overline{\texttt {Y}}},\theta _\star \}} O(r-r_-)\,\omega ^\sharp _\dagger \otimes X_\ddagger \\{} & {} +O(r-r_-)\,\overline{Y}\otimes \frac{1}{\sin \theta }\partial _{\phi _\star } +O(r-r_-)\,\frac{1}{\sin \theta }\partial _{\phi _\star }\otimes \overline{Y}\\{} & {} +O(r-r_-)\,\overline{T}\otimes \frac{1}{\sin \theta }\partial _{\phi _\star } +O(r-r_-)\,\frac{1}{\sin \theta }\partial _{\phi _\star }\otimes \overline{T}\\{} & {} +O(r-r_-)\,\partial _{\theta _\star }\otimes \frac{1}{\sin \theta }\partial _{\phi _\star } +O(r-r_-)\,\frac{1}{\sin \theta }\partial _{\phi _\star }\otimes \partial _{\theta _\star }\\{} & {} +O(r-r_-)\frac{1}{\sin ^2\theta }\partial _{\phi _\star }\otimes \partial _{\phi _\star } \end{aligned}$$

because \(\omega _{\phi _\star }^\sharp =\partial ^{\phi _\star }\) behaves like \({g \hspace{-4.83322pt}/}^{\phi _\star \phi _\star }\partial _{\phi _\star }\), which in turn behaves like \(\frac{1}{\sin ^2\theta }\partial _{\phi _\star }\).

3.1.3 Currents

3.1.3.1 The energy momentum tensor of a massless scalar field and the vector current

The energy momentum tensor is given by

$$\begin{aligned} T_{\mu \nu }=\partial _\mu \psi \partial _\nu \psi -\,\frac{1}{2}g_{\mu \nu }\partial ^\alpha \psi \,\partial _\alpha \psi , \end{aligned}$$

and one readily checks that

$$\begin{aligned} \partial ^\alpha \psi \,\partial _\alpha \psi& = -\,\frac{1}{\Omega ^2}(\partial _u\psi )( \partial _v\psi +b^{\phi _{\star }}\partial _{\phi _{\star }}\psi ) +|{\nabla \hspace{-7.22214pt}/ \,}\psi |_{g \hspace{-4.83322pt}/}^2\\& = -\,\frac{1}{\Omega ^2}(\partial _u\psi )( \partial _v\psi +b^{\phi _{\star }}\partial _{\phi _{\star }}\psi ) +{g \hspace{-4.83322pt}/}^{\theta _\star \theta _\star }(\partial _{\theta _\star }\psi )^2\\{} & {} +2{g \hspace{-4.83322pt}/}^{\theta _\star \phi _\star }(\partial _{\theta _\star }\psi )(\partial _{\phi _\star }\psi ) +{g \hspace{-4.83322pt}/}^{\phi _\star \phi _\star }(\partial _{\phi _\star }\psi )^2. \end{aligned}$$

The energy-momentum tensor is written as

$$\begin{aligned} T=& {} \frac{1}{4} \left( \partial _{u}\psi \right) ^2\omega _{{\tiny \overline{\texttt {T}}}}\otimes \omega _{{\tiny \overline{\texttt {T}}}}+ \frac{1}{2}|{\nabla \hspace{-7.22214pt}/ \,}\psi |_{g \hspace{-4.83322pt}/}^2\,(\omega _{{\tiny \overline{\texttt {T}}}}\otimes \omega _{{\tiny \overline{\texttt {Y}}}}+\omega _{{\tiny \overline{\texttt {Y}}}}\otimes \omega _{{\tiny \overline{\texttt {T}}}})\\{} & {} + \frac{1}{\Omega ^4_{{\tiny \mathcal{C}\mathcal{H}^+}}}\left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\psi +b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\psi \right) ^2\omega _{{\tiny \overline{\texttt {Y}}}}\otimes \omega _{{\tiny \overline{\texttt {Y}}}}\\{} & {} +\frac{1}{2} \left( \partial _{u}\psi \right) (\partial _{\theta _\star }\psi ) (\omega _{{\tiny \overline{\texttt {T}}}}\otimes \omega _{\theta _\star }+\omega _{\theta _\star }\otimes \omega _{{\tiny \overline{\texttt {T}}}}) +\frac{1}{2}\left( \partial _{u}\psi \right) (\partial _{\phi _\star }\psi ) (\omega _{{\tiny \overline{\texttt {T}}}}\otimes \omega _{\phi _\star }+\omega _{\phi _\star }\otimes \omega _{{\tiny \overline{\texttt {T}}}})\\{} & {} +\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\psi +b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\psi \right) (\partial _{\theta _\star }\psi )(\omega _{{\tiny \overline{\texttt {Y}}}}\otimes \omega _{\theta _\star }+\omega _{\theta _\star }\otimes \omega _{{\tiny \overline{\texttt {Y}}}})\\{} & {} +\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}} \left( \partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\psi +b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\psi \right) (\partial _{\phi _\star }\psi )(\omega _{{\tiny \overline{\texttt {Y}}}}\otimes \omega _{\phi _\star }+\omega _{\phi _\star }\otimes \omega _{{\tiny \overline{\texttt {Y}}}})\\{} & {} +\left( (\partial _{\theta _\star }\psi )^2-\,\frac{{g \hspace{-4.83322pt}/}_{\theta _{\star },\theta _{\star }}}{2}\left( -\,\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}(\partial _u\psi ) (\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\psi +b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _{\star }}\psi ) +|{\nabla \hspace{-7.22214pt}/ \,}\psi |_{g \hspace{-4.83322pt}/}^2\right) \right) \omega _{\theta _\star }\otimes \omega _{\theta _\star }\\{} & {} +\left( (\partial _{\theta _\star }\psi )(\partial _{\phi _\star }\psi )-\,\frac{{g \hspace{-4.83322pt}/}_{\theta _{\star },\phi _{\star }}}{2}\left( -\,\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}(\partial _u\psi ) (\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\psi +b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _{\star }}\psi ) +|{\nabla \hspace{-7.22214pt}/ \,}\psi |_{g \hspace{-4.83322pt}/}^2\right) \right) (\omega _{\theta _\star }\otimes \omega _{\phi _\star }+\omega _{\phi _\star }\otimes \omega _{\theta _\star })\\{} & {} +\left( (\partial _{\phi _\star }\psi )^2-\,\frac{{g \hspace{-4.83322pt}/}_{\phi _{\star },\phi _{\star }}}{2}\left( -\,\frac{1}{\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}}(\partial _u\psi ) (\partial _{v_{{\tiny \mathcal{C}\mathcal{H}^+}}}\psi +b_{{\tiny \mathcal{C}\mathcal{H}^+}}^{\phi _\star }\partial _{\phi _{\star }}\psi ) +|{\nabla \hspace{-7.22214pt}/ \,}\psi |_{g \hspace{-4.83322pt}/}^2\right) \right) \omega _{\phi _\star }\otimes \omega _{\phi _\star }. \end{aligned}$$

The vector currents associated to the blue-shift and red-shift vector fields are

$$\begin{aligned} J^{N_b}_\mu =T_{\mu \nu }N_b^\nu \qquad \hbox {and}\qquad J^{N_r}_\mu =T_{\mu \nu }N_r^\nu , \end{aligned}$$

respectively.

3.1.3.2 The scalar current associated to the blue-shift vector field

The scalar current associated to \(N_b\) is \(K^{N_b}=T_{\mu \nu }\nabla ^\mu N_b^\nu\). Since Z is a Killing vector field this is equal to \(K^Y\), which is

$$\begin{aligned} K^Y& = T_{\mu \nu }\nabla ^\mu Y^\nu \\& = \kappa _-T(\overline{Y},\overline{Y}) -2T(\overline{Y},a^{\theta _\star }\partial _{\theta _\star }) -2T(\overline{Y},a^{\phi _\star }\partial _{\phi _\star })\\{} & {} +\iota T(\overline{T},\overline{Y})+\iota T(\overline{T},\overline{T})\\{} & {} +h^{\theta _\star \theta _\star }T(\partial _{\theta _\star },\partial _{\theta _\star }) +h^{\theta _\star \phi _\star }T(\partial _{\theta _\star },\partial _{\phi _\star }) +h^{\phi _\star \theta _\star }T(\partial _{\phi _\star },\partial _{\theta _\star }) +h^{\phi _\star \phi _\star }T(\partial _{\phi _\star },\partial _{\phi _\star })\\{} & {} +O(r-r_-)(T(\overline{T},\overline{T})+T(\overline{T},\overline{Y})+T(\overline{Y},\overline{Y})). \end{aligned}$$

We estimate \(K^Y\). Suppose we are given \(\delta \in \bigl (0,\frac{1}{3}\bigr )\). Let \(\overline{c}=2(-\kappa _-)\delta\). Choose \(\iota =c+\overline{c}+1\), where \(c=c_\delta\) is the constant below (which is independent of Y). There exists \(r_0>r_-\) such that \(r\in (r_-,r_0)\) implies that

$$\begin{aligned} K^Y\ge & {} \kappa _-T(\overline{Y},\overline{Y}) +\iota T(\overline{T},\overline{Y})+\iota T(\overline{T},\overline{T})\nonumber \\{} & {} -cT(\overline{T},\overline{T})-cT(\overline{T},\overline{Y})-(-\kappa _-)(\delta /2) T(\overline{Y},\overline{Y})\nonumber \\{} & {} +O(r-r_-)(T(\overline{T},\overline{T})+T(\overline{T},\overline{Y})+T(\overline{Y},\overline{Y}))\nonumber \\\ge & {} \kappa _-(1+\delta )T(\overline{Y},\overline{Y})+\overline{c}(T(\overline{T},\overline{T})+ T(\overline{T},\overline{Y})). \end{aligned}$$
(101)

We have used the fact that

$$\begin{aligned} {g \hspace{-4.83322pt}/}^{\phi _\star \phi _\star }(\partial _{\phi _\star }\psi )^2\le |{\nabla \hspace{-7.22214pt}/ \,}\psi |^2_{g \hspace{-4.83322pt}/}. \end{aligned}$$
3.1.3.3 Inequalities relating the currents

The blue-shift vector field satisfies

Lemma 3.3

Let \(0<\delta <\frac{1}{3}\). If \((r_\star )_0\) is chosen sufficiently large (see Remark 2.1), then

$$\begin{aligned} \int \limits _{{\mathop {\tiny {v>(r_\star )_0-u}}\limits ^{u_0<u<u_1}}} K^{N_b}\textrm{dVol}\ge 2\kappa _-(1+3\delta )\int _{[u_0,u_1]}\left( \int _{v>(r_\star )_0-u} J^{N_b}_\mu n_{\mathcal{C}_u}^\mu \textrm{dVol}_{{\mathcal{C}_u}}\right) {\,\textrm{d}}u. \end{aligned}$$
(102)

Proof

Using (27) and (101), we obtain

$$\begin{aligned} \int \limits _{{\mathop {\tiny { v>(r_\star )_0-u}}\limits ^{u_0<u<u_1}}} K^N\textrm{dVol}\ge \int \limits _{{\mathop {\tiny { v>(r_\star )_0-u}}\limits ^{u_0<u<u_1}}} \Bigl(2\kappa _-(1+\delta )T(\overline{Y},\overline{Y})+ 2\overline{c}( T(\overline{T},\overline{T})+ T(\overline{T},\overline{Y}))\Bigr)\Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}\frac{L\sin \theta }{\Xi } {\,\textrm{d}}u{\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}{\,\textrm{d}}\theta _\star {\,\textrm{d}}\phi _\star . \end{aligned}$$

Using (26), we see that

$$\begin{aligned} T\left( N_b,\frac{n_{\mathcal{C}_u}}{\Omega ^2}\right)& = T(N_b,\overline{Y})\\& = T(\overline{T}+\overline{Y},\overline{Y})+O(r-r_-) \left( T(\overline{T},\overline{Y})+T(\overline{Y},\overline{Y}) +T(\partial _{\theta _\star },\overline{Y})+T(\partial _{\phi _\star },\overline{Y}) \right) . \end{aligned}$$

For each \(0<\delta <\frac{1}{3}\), there exists \(r_0>r_-\) such that \(r\in (r_-,r_0)\) implies that

$$\begin{aligned} \int \limits _{v>(r_\star )_0-u} T\left( N_b,\frac{n_{\mathcal{C}_u}}{\Omega ^2}\right) \Omega ^2\textrm{dVol}_{{\mathcal{C}_u}}\ge \int \limits _{v>(r_\star )_0-u} \left( (1-\delta )\left( T(\overline{T},\overline{Y})+ T(\overline{Y},\overline{Y})\right) -\delta T(\overline{T},\overline{T}) \right) \Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}\frac{L\sin \theta }{\Xi } {\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}{\,\textrm{d}}\theta _\star {\,\textrm{d}}\phi _\star . \end{aligned}$$

Multiplying both sides by \(2\kappa _-(1+\delta )/(1-\delta )\), we get

$$\begin{aligned}{} & {} 2\kappa _-(1+\delta )\int \limits _{v>(r_\star )_0-u} \left( \left( T(\overline{T},\overline{Y})+ T(\overline{Y},\overline{Y})\right) -\, \frac{\delta }{1-\delta } T(\overline{T},\overline{T})\right) \Omega ^2_{{\tiny \mathcal{C}\mathcal{H}^+}}\frac{L\sin \theta }{\Xi } {\,\textrm{d}}v_{{\tiny \mathcal{C}\mathcal{H}^+}}{\,\textrm{d}}\theta _\star {\,\textrm{d}}\phi _\star \\{} & {} \qquad \ge 2\kappa _-\,\frac{1+\delta }{1-\delta }\int \limits _{v>(r_\star )_0-u} T\left( N_b,\frac{n_{\mathcal{C}_u}}{\Omega ^2}\right) \Omega ^2\textrm{dVol}_{{\mathcal{C}_u}}. \end{aligned}$$

This implies (102) because \(\frac{1+\delta }{1-\delta }<1+3\delta\) and \(-2\kappa _-\delta \frac{1+\delta }{1-\delta }<-4\kappa _-\delta =2\overline{c}\). \(\square\)

Similarly to (102), the red-shift vector field satisfies

Lemma 3.4

Let \(\delta >0\). For \((r_\star )_0\) sufficiently negative (see Remark 2.1), we have

$$\begin{aligned} \int _{{\mathop {\tiny { u\!\!<\!\!(r_\star )_0\!\!-\!\!v}}\limits ^{v_0<v<v_1}}} K^{N_r}\textrm{dVol}\ge 2\kappa _+(1-\delta )\int _{[v_0,v_1]}\left( \int _{u<(r_\star )_0-v} J^{N_r}_\mu n_{\underline{\mathcal{C}}_v}^\mu \textrm{dVol}_{\underline{\mathcal{C}}_v}\right) {\,\textrm{d}}v. \end{aligned}$$
(103)

Proof

Work in the frame \((\overline{V},\underline{T},\partial _{\theta _\star },\partial _{\phi _\star })\), with

$$\begin{aligned} \overline{V}={\textstyle \frac{1}{\Omega ^2_{{\tiny \mathcal{H}^+}}}}\partial _{u_{\mathcal{H}^+}}, \qquad \underline{T}=\frac{1}{2}\left( \partial _{v_{\mathcal{H}^+}}+b_{{\tiny \mathcal{H}^+}}^{\phi _\star }\partial _{\phi _\star }\right) . \end{aligned}$$

\(\square\)

3.2 Energy estimates

We are interested in solutions of the wave equation which are regular up to, and including, \(\mathcal{H}^+\), and which have compact support on \(\mathcal{H}^+\), i.e. we are interested in functions belonging to the space

$$\begin{aligned} \mathcal{F}:=\left\{ \psi \in C^\infty (\mathcal{M}\cup \mathcal{H}^+):\Box _g\psi =0\ \hbox {and there exists}\ v_0\in {{\mathbb {R}}}\ \hbox {such that}\ \psi |_{\mathcal{H}^+\cap \{v\ge v_0\}}=0\right\} . \end{aligned}$$

The following theorem, established by Sbierski in his thesis [29] for Reissner–Nordström and Kerr black holes, applies to Kerr–Newman–de Sitter black holes.

Theorem 3.5

Let \(2\kappa _+>-\kappa _-\) and \(\psi \in \mathcal{F}\). Then, for any \(u_0\), \(v_0\in {{\mathbb {R}}}\), we have

$$\begin{aligned}{} & {} \int \limits _{{\mathop {{\tiny { (-\infty ,u_0]}}}\limits ^{\mathcal{C}\mathcal{H}^+}}} J^{N_b}_\mu n^\mu _{\mathcal{C}\mathcal{H}^+}\textrm{dVol}_{\mathcal{C}\mathcal{H}^+}+ \int \limits _{{\mathop {{\tiny {[v_0,\infty )}}}\limits ^{\mathcal{C}_{u_0}}}} J^{N_b}_\mu n^\mu _{\mathcal{C}_{u_0}}\textrm{dVol}_{{\mathcal{C}_{u_0}}} <+\infty . \end{aligned}$$

Proof

We sketch the proof and refer to [29] for further details.

(a) Estimates in the red-shift region. For \(\kappa <\kappa _+\), define \(\underline{N}_r=e^{2\kappa v}N_r\). Choose the \(\delta\) in (103) such that \(\kappa <\kappa _+(1-\delta )\). Since \(K^{\underline{N}_r}\ge 0\), the Divergence Theorem implies that we have

$$\begin{aligned} \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]\cap [v_0,v_1]}}}\limits ^{\Sigma _{(r_\star )_0}}}} J^{\underline{N}_r}_\mu n^\mu _{\Sigma _{(r_\star )_0}}\textrm{dVol}_{\Sigma _{(r_\star )_0}} \le \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]}}}\limits ^{\underline{\mathcal{C}}_{v_0}}}} J^{\underline{N}_r}_\mu n^\mu _{\underline{\mathcal{C}}_{v_0}}\textrm{dVol}_{\underline{\mathcal{C}}_{v_0}} =e^{2\kappa v_0}\int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]}}}\limits ^{\underline{\mathcal{C}}_{v_0}}}} J^{N_r}_\mu n^\mu _{\underline{\mathcal{C}}_{v_0}}\textrm{dVol}_{\underline{\mathcal{C}}_{v_0}}. \end{aligned}$$

The left-hand side can be bounded below by

$$\begin{aligned} \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]\cap [v_0,v_1]}}}\limits ^{\Sigma _{(r_\star )_0}}}} J^{\underline{N}_r}_\mu n^\mu _{\Sigma _{(r_\star )_0}}\textrm{dVol}_{\Sigma _{(r_\star )_0}}\ge e^{2\kappa (r_\star )_0}e^{-2\kappa u(u_{\mathcal{H}^+})} \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]\cap [v_0,v_1]}}}\limits ^{\Sigma _{(r_\star )_0}}}} J^{N_r}_\mu n^\mu _{\Sigma _{(r_\star )_0}}\textrm{dVol}_{\Sigma _{(r_\star )_0}}, \end{aligned}$$

where \(u(u_{\mathcal{H}^+})\) denotes the u corresponding to \(u_{\mathcal{H}^+}\). As \(\psi\) is a regular function on \(\mathcal{H}^+\), we have

$$\begin{aligned}{} & {} \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]}}}\limits ^{\underline{\mathcal{C}}_{v_0}}}}J^{N_r}_\mu n^\mu _{\underline{\mathcal{C}}_{v_0}}\textrm{dVol}_{\underline{\mathcal{C}}_{v_0}}\\{} & {} \qquad \qquad \le C \int _0^{u_{\mathcal{H}^+}}\left( \left( \frac{(\partial _{\tilde{u}_{\mathcal{H}^+}}\psi )^2}{\Omega ^4_{{\tiny \mathcal{H}^+}}} +\frac{1}{2}|{\nabla \hspace{-7.22214pt}/ \,}\psi |^2_{g \hspace{-4.83322pt}/}\right) \Omega ^2_{{\tiny \mathcal{H}^+}}\frac{L\sin \theta }{\Xi } \right) (\,\cdot ,v_0,\,\cdot ,\,\cdot \,)\,{\,\textrm{d}}\tilde{u}_{\mathcal{H}^+}{\,\textrm{d}}\theta _\star {\,\textrm{d}}\phi _\star \\{} & {} \qquad \qquad \le Cu_{\mathcal{H}^+}=Ce^{2\kappa _+u(u_{\mathcal{H}^+})}. \end{aligned}$$

We conclude that

$$\begin{aligned} \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]\cap [v_0,v_1]}}}\limits ^{\Sigma _{(r_\star )_0}}}} J^{N_r}_\mu n^\mu _{\Sigma _{(r_\star )_0}}\textrm{dVol}_{\Sigma _{(r_\star )_0}}\le Ce^{4\kappa u(u_{\mathcal{H}^+})}. \end{aligned}$$

(b) Estimates in the no-shift region. We recall [29, Lemma 4.5.6]: Given \((r_\star )_1>(r_\star )_0\) and a smooth future directed timelike time invariant vector field N, there exists a constant \(C>0\) such that

$$\begin{aligned} \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]}}}\limits ^{\Sigma _{(r_\star )_1}}}}J^N_\mu n^\mu _{\Sigma _{(r_\star )_1}} \textrm{dVol}_{\Sigma _{(r_\star )_1}} \le C \int \limits _{{\mathop {{\tiny { [0,u_{\mathcal{H}^+}]}}}\limits ^{\Sigma _{(r_\star )_0}}}} J^N_\mu n^\mu _{\Sigma _{(r_\star )_0}} \textrm{dVol}_{\Sigma _{(r_\star )_0}} \end{aligned}$$
(104)

holds for all solutions of the wave equation.

(c) Estimates in the blue-shift region. We also recall [7, Lemma 4.5]: Let \(f:[t_0,\infty [\rightarrow {{\mathbb {R}}}\) and assume that for some \(\alpha _1, C>0\), and for all \(t\ge t_0\),

$$\begin{aligned} \int _t^\infty f(s)\,ds\le Ce^{-\alpha _1 t}. \end{aligned}$$

Then, for all \(0<\alpha _2<\alpha _1\) and \(t\ge t_0\), we have

$$\begin{aligned} \int _t^\infty e^{\alpha _2s} f(s)\,ds\le Ce^{-(\alpha _1-\alpha _2)t}. \end{aligned}$$
(105)

By assumption \(2\kappa _+>-\kappa _-\). If we choose \(\kappa <\kappa _+\) sufficiently close to \(\kappa _+\) and and \(\overline{\kappa }<\kappa _-\) sufficiently close to \(\kappa _-\), then \(2\kappa >-\overline{\kappa }\). Using (104), (105) and the fact that

$$\begin{aligned} \int \limits _{{\mathop {{\tiny { (-\infty ,u]}}}\limits ^{\Sigma _{(r_\star )_1}}}} J^{N_r}_\mu n^\mu _{\Sigma _{(r_\star )_1}} \textrm{dVol}_{\Sigma _{(r_\star )_1}} \quad \hbox {and}\quad \int \limits _{{\mathop {{\tiny { (-\infty ,u]}}}\limits ^{\Sigma _{(r_\star )_1}}}} J^{N_b}_\mu n^\mu _{\Sigma _{(r_\star )_1}} \textrm{dVol}_{\Sigma _{(r_\star )_1}} \end{aligned}$$

are comparable, we conclude that

$$\begin{aligned} \int \limits _{{\mathop {{\tiny { (-\infty ,u_0]\cap \left( -\infty ,\left( v_{{\tiny \mathcal{C}\mathcal{H}^+}}\right) _0\right] }}}\limits ^{\Sigma _{(r_\star )_1}}}} e^{-2\overline{\kappa } u}J^{N_b}_\mu n^\mu _{\Sigma _{(r_\star )_1}}\textrm{dVol}_{\Sigma _{(r_\star )_1}}<+\infty . \end{aligned}$$

Define \(\underline{N}_b=e^{-2\overline{\kappa }u}N_b\). Choose the \(\delta\) in (102) such that \(\overline{\kappa }<\kappa _-(1+3\delta )\). Since \(K^{\underline{N}_b}\ge 0\), the Divergence Theorem implies that

$$\begin{aligned}{} & {} \int \limits _{{\mathop {{\tiny { (-\infty ,u_0]\cap \left( -\infty ,\left( v_{{\tiny \mathcal{C}\mathcal{H}^+}}\right) _0\right] }}}\limits ^{\Sigma _{(r_\star )_2}}}} e^{-2\overline{\kappa } u}J^{N_b}_\mu n^\mu _{\Sigma _{(r_\star )_2}}\textrm{dVol}_{\Sigma _{(r_\star )_2}}+ \int \limits _{{\mathop {{\tiny { (r_\star )_1-u_0<v\left( v_{{\tiny \mathcal{C}\mathcal{H}^+}}\right) < (r_\star )_2-u_0}}}\limits ^{\mathcal{C}_{u_0}}}} e^{-2\overline{\kappa } u_0} J^{N_b}_\mu n^\mu _{\mathcal{C}_{u_0}}\textrm{dVol}_{{\mathcal{C}_{u_0}}}\\{} & {} \qquad \qquad \qquad \le \int \limits _{{\mathop {{\tiny { (-\infty ,u_0]\cap \left( -\infty ,\left( v_{{\tiny \mathcal{C}\mathcal{H}^+}}\right) _0\right] }}}\limits ^{\Sigma _{(r_\star )_1}}}} e^{-2\overline{\kappa } u}J^{N_b}_\mu n^\mu _{\Sigma _{(r_\star )_1}}\textrm{dVol}_{\Sigma _{(r_\star )_1}}. \end{aligned}$$

This finishes the proof. \(\square\)

This theorem shows that for KNdS black holes, with surface gravities satisfying \(2\kappa _+>-\kappa _-\), the energy of a wave with compact support on \(\mathcal{H}^+\) has finite energy along a null hypersurface intersecting the Cauchy horizon. This suggests that there exists a set of of black hole parameters for which compactly supported perturbations do not lead to mass inflation and to instability of the Cauchy horizon. It is one more step towards understanding the conjectured instability of black hole interiors in the context of KNdS spacetimes.