On the linear stability of the extreme Kerr black hole under axially symmetric perturbations

Recently there has been considerable progress on the long-standing and central open problem of black hole stability in general relativity (see the review articles [14, 15] and references therein). The following three aspects of this problem motivated the present work.

(i) Non-modal stability of linear gravitational perturbations: the non-modal stability of linear gravitational perturbations for the Kerr black hole still remains unsolved. The works of Regge and Wheeler [39], Zerilli [46, 47] and Moncrief [36] determined the modal linear stability of gravitational perturbations for the Schwarzschild black hole by ruling out exponential growth in time for every individual mode. The modal stability for the Kerr black hole was proved by Whiting [45] using the Teukolsky equation. However, modal stability is not enough to exclude that general linear perturbations grow unbounded in time (see, for example, the discussion in [15, 43]). The study of black hole non-modal stability was initiated by Kay and Wald in [32, 43]. They prove that solutions of the linear wave equation on a Schwarzschild black hole background remain bounded by a constant for all time. An important ingredient in this proof is the use of conserved energies to control the norm of the solution. The analog of the Kay–Wald theorem on a large class of backgrounds, which includes the slow rotating Kerr black hole, was first proved by Dafermos and Rodniaski [16] and then, independently, in the special case of slow rotating Kerr by Andersson and Blue [1]. In [15] Dafermos and Rodniaski provide the essential elements of the proof of this theorem for the general subextremal Kerr black hole. Recently, this problem was finally solved in [17]. For a complete list of references with important related works on this subject, see the review articles [14, 15, 28]. All these results concern the wave equation. For gravitational perturbations, the only non-modal stability result was given very recently by Dotti [26] for the Schwarzschild black hole. There are, so far, no results regarding the non-modal stability of the Kerr black hole under linear gravitational perturbations.

(ii) Stability and instability of extreme black holes: extreme black holes are relevant because they lie on the boundary between black holes and naked singularities, and hence it is expected that their study shed light on the cosmic censorship conjecture. Recently, Aretakis discovered certain instabilities for extreme black holes [3, 4]. These instabilities concern transverse derivatives of the field at the horizon: a conservation law ensures that the first transverse derivative of the field on the event horizon generically does not decay; this implies that the second transverse derivative of the field generically grows with time on the horizon. These instabilities were discovered first for the scalar wave equation on the extreme Reissner–Nordström black hole; a similar result also holds for the extreme Kerr black hole [5, 6]. These works were extended in several directions: for generic extreme black holes and linear gravitational perturbations [35], for certain higher dimensional extreme vacuum black holes [37], for massive scalar field and for coupled linearized gravitational and electromagnetic perturbations [34], and for a test scalar field with a non-linear self-interaction in the extreme Kerr geometry [7]. An interesting relation between these instabilities and the Newman–Penrose constants was pointed out by Bizon and Friedrich [9]. This relation was also independently observed by Lucietti, Murata, Reall, and Tanahashi [34]. Finally, a numerical study of non-linear evolution of this instability for spherically symmetric perturbations of an extreme Reissner–Nordström black hole was performed by Murata et al in [38].

An important question regarding the dynamical behavior of extremal black holes is whether a non-extremal black hole can later evolve to an extremal one. In [40] Reiris proved that there exists arbitrary small perturbations of the extreme black hole initial data that cannot decay in time into any extreme black hole. On the other hand, in [38] fine-tuned initial data are numerically constructed, which settle to an extreme Reissner–Nordström black hole. There is no contradiction between these two results since they apply to different kinds of data. It is interesting to note that the construction in [40] relies on geometrical inequalities between area and charges on trapped surfaces (see [21] and references therein); in contrast, in the spacetime considered in [38], there are no trapped surfaces.

The discussion noted earlier concerns the instability of extreme black holes. However, there are also stability results for this class of black holes. The most relevant of them is that the solutions of the wave equations remain pointwise bounded in the black hole exterior region [3] (see also [22]).

(iii) Non-linear stability: the problem of the black hole non-linear stability remains largely open (see the discussion in [15] and references therein). The linear studies previously discussed are expected to provide insight into the non-linear problem. However, this will be possible only if they rely on techniques that can be suitably extended to the non-linear regime. One of the most important of these techniques is the energy estimates.

The main result of this article is the following.

For axially symmetric linear gravitational perturbation of the extreme Kerr black hole there exists an energy that is positive, definite, and conserved.

A precise version of this statement is given in theorem 4.1. In the following, we discuss the relation of this result with the points (i), (ii), and (iii) discussed earlier.

(i) The conserved energy for the linear perturbation has a similar structure as the energy of the wave equation: it is an integral over a spacelike surface of terms that involves squares of first derivatives of the perturbations. This energy is related to the second-order expansion of the ADM mass. However, it is important to stress that the positiveness of this energy cannot be easily deduced from the positiveness of the ADM mass. In fact, as we will see, this result is proved as a consequence of highly non-trivial identities. It is also important to emphasize that this energy is positive inside the ergosphere too.

The energy expression and its conservation do not require any mode expansion of the fields. The existence of this conserved quantity provides a basic non-modal stability criteria for axially symmetric linear perturbation of the extreme Kerr black hole. Since the equations are linear and the coefficients of them do not depend on time, it is possible to construct an infinitely number of higher order conserved energies. We expect that these higher order energies can be used to prove the pointwise boundedness of the solution, in a similar fashion as in [22]. In that reference the pointwise boundedness of solutions of the wave equation on the extreme Reissner–Nordström black hole was proved using only higher order energy estimates. But, up to now, we have not been able to extend this result to the present context. However, in the particular case of the Minkowski background, we prove a pointwise bound for the linear perturbations in a remarkably simple way. Comparing with the Minkowski case, there are two main difficulties to obtaining pointwise estimates from the energy in the Kerr case: first, the equations for the norm and the twist are coupled, and hence it is not possible to separate them as in the Minkowski case. Second, the coefficient of the equations are singular at the horizon, and hence we can not use standard Sobolev estimates.

This conserved energy is closely related to the energy studied by Hollands and Wald [31] (see also [33]). We expect that the techniques used here to prove positiveness should also be useful in that context. Also, the boundary conditions at the horizon proposed in [31] are likely to be useful to generalize our results to the non-extreme case.

(ii) The existence of this conserved energy and its related stability criteria are not in contradiction with Aretakis instabilities. The situation is very similar to the one discussed in [22] for the case of the wave equation: the energy is only defined in the black hole exterior region, and it does not control any transverse derivative at the horizon.

(iii) As we pointed out earlier, the energy used here is related to the ADM mass, which is also conserved in the non-linear regime (see the discussion in [24]). That is, the energy estimates used here are very likely to be useful in the non-linear case.

The plan of the article is as follows. The expression of the conserved energy arises naturally in a particular gauge for the Einstein equation: the maximal-isothermal gauge. We review this gauge in section 2. In that section we also present the linearized equations on a class of stationary backgrounds. In section 3 we study the particular case of the Minkowski background, where we prove that the solutions are pointwise bounded in terms of a constant that depends only on the conserved energy (see theorem 3.1). In section 4 we study the extreme Kerr background and we prove the main result of this article given by theorem 4.1. Finally, in the appendices we write the Kerr solution in the maximal-isothermal gauge and we also prove a Sobolev-like estimate needed in the proof of theorem 3.1.

In axial symmetry, the maximal-isothermal gauge has the important property that the total ADM mass can be written as a positive definite integral on the spacelike hypersurfaces of the foliation and the integral is constant along the evolution [19]. The conserved energy for the linear perturbations will be obtained as an appropriate second order expansion of this integral. In this section we first review the full Einstein equations in this gauge in section 2.1, and then in section 2.2 we perform the linearization on a class of stationary backgrounds that includes the Kerr black hole. On this class of backgrounds the linearized equations in this gauge have a remarkably simple form.

2.1. Einstein equations

Einstein equations in the maximal-isothermal gauge were studied, with slight variations, in several works [11, 24, 29, 41]. In this section we review these equations; we closely follow [24].

In axial symmetry, it is possible to perform a symmetrical reduction of Einstein equations to obtain a set of geometrical equations in the three-dimensional quotient manifold in terms of a Lorenzian three-dimensional metric. See [24] for the details. In appendix A we explicitly perform this reduction for the Kerr metric.

On the three-dimensional quotient manifold we take a foliation of spacelike surfaces. The intrinsic metric on the slices of the foliation is denoted by q_AB and the extrinsic curvature by ${{\chi }_{AB}}$. Here the indices $A,B\cdots$ are two-dimensional.

The maximal-isothermal gauge and its associated cylindrical coordinates $(t,\rho ,z)$ are defined by the following two conditions. For the the lapse, denoted by α, we impose the maximal condition on the 2-surfaces t = constant. That is, the trace χ of the extrinsic curvature vanishes

$$\begin{eqnarray}&&\chi ={{q}^{AB}}{{\chi }_{AB}}=0.\end{eqnarray} \tag{ 1 }$$

The shift, denoted by ${{\beta }^{A}}$, is fixed by the requirement that the intrinsic metric q_AB has the following form:

$$\begin{eqnarray}&&{{q}_{AB}}={{{\rm e}}^{2u}}{{\delta }_{AB}},\end{eqnarray} \tag{ 2 }$$

where ${{\delta }_{AB}}$ is the fixed flat metric

$$\begin{eqnarray}&&\delta ={\rm d}{{\rho }^{2}}+{\rm d}{{z}^{2}}.\end{eqnarray} \tag{ 3 }$$

For our purposes, the relevant geometries for the two-dimensional spacelike surfaces are the half-plane $\mathbb{R}_{+}^{2}$ (defined by $-\infty \lt z\lt \infty$, $0\leqslant \rho \lt \infty$) for the Minkowski case, or $\mathbb{R}_{+}^{2}\backslash \{0\}$ for the black hole case. In that case, the origin will represent an extra asymptotic end. For both cases, the axis of symmetry is defined by $\rho =0$.

The dynamical degree of freedom of the gravitational field is encoded in two geometrical scalars η and ω, the square of the norm, and the twist of the axial Killing vector, respectively. Due to the behavior at the axis, instead of η, α, and u, it is often convenient to work with the auxiliary function σ, $\bar{\alpha }$, and q defined by

$$\begin{eqnarray}&&\eta ={{\rho }^{2}}{{{\rm e}}^{\sigma }},\qquad \quad \alpha =\rho \bar{\alpha },\qquad \quad u={\rm ln} \rho +\sigma +q.\end{eqnarray} \tag{ 4 }$$

To write the equations, we will make use of the following differential operators: the two-dimensional Laplacian $\Delta$ defined by

$$\begin{eqnarray}&&\Delta q=\partial _{\rho }^{2}q+\partial _{z}^{2}q,\end{eqnarray} \tag{ 5 }$$

and the operator $^{(3)}\Delta$ defined as

$$\begin{eqnarray}&&^{(3)}\Delta \sigma =\Delta \sigma +\frac{{{\partial }_{\rho }}\sigma }{\rho }.\end{eqnarray} \tag{ 6 }$$

This operator, which appears frequently in the rest of the article, is the flat Laplace operator in three dimensions written in cylindrical coordinates and acting on axially symmetric functions. The conformal Killing operator $\mathcal{L}$ acting on a vector ${{\beta }_{A}}$ is defined by

$$\begin{eqnarray}&&{{(\mathcal{L}\beta )}_{AB}}={{\partial }_{A}}{{\beta }_{B}}+{{\partial }_{B}}{{\beta }_{A}}-{{\delta }_{AB}}{{\partial }_{C}}{{\beta }^{C}}.\end{eqnarray} \tag{ 7 }$$

In these equations ∂ denotes partial derivatives with respect to the space coordinates $(\rho ,z)$ and all the indices are moved with the flat metric ${{\delta }_{AB}}$. We denote by a dot the partial derivative with respect to t, and we define the prime operator as

$$\begin{eqnarray}&&\eta ^{\prime} =\frac{1}{\alpha }\left( \dot{\eta }-{{\beta }^{A}}{{\partial }_{A}}\eta \right).\end{eqnarray} \tag{ 8 }$$

Einstein equations in the maximal-isothermal gauge are divided into three groups: evolution equations, constraint equations, and gauge equations. The evolution equations are further divided into two groups: evolution equations for the dynamical degree of freedom $(\sigma ,\omega )$ and evolution equations for the metric q_AB and second fundamental form ${{\chi }_{AB}}$. Due to the axial symmetry, these equations are not independent (see the discussion in [41]). For example, the constraint equations are essentially equivalent to the evolution equations for the metric and second fundamental form. In particular, in this article we will not make use of the evolution equations for the metric and second fundamental form; we will always use instead a time derivative of the constraint equations.

We write the following equations; for the deduction of them, see [24]. We divide them in the three groups discussed earlier. In the next sections the linearization of these equations on a different background is performed; for the sake of clarity, we will always group them in the same way.

Evolution equations: the evolution equations for σ and ω are given by³

$$\begin{eqnarray}&&-{{{\rm e}}^{2u}}\sigma ^{\prime\prime} {{+}^{(3)}}\Delta \sigma +{{\partial }_{A}}\sigma \frac{{{\partial }^{A}}\bar{\alpha }}{{\bar{\alpha }}}-2{{{\rm e}}^{2u}}\left( {\rm log} \rho \right)^{\prime\prime} +2\frac{{{\partial }_{\rho }}\bar{\alpha }}{\bar{\alpha }\rho }=\frac{\left( {{{\rm e}}^{2u}}\omega {{^{\prime} }^{2}}-|\partial \omega {{|}^{2}} \right)}{{{\eta }^{2}}},\end{eqnarray} \tag{ 9 }$$

$$\begin{eqnarray}&&-{{{\rm e}}^{2u}}\omega ^{\prime\prime} {{+}^{(3)}}\Delta \omega +{{\partial }_{A}}\omega \frac{{{\partial }^{A}}\bar{\alpha }}{{\bar{\alpha }}}=\frac{2\left( {{\partial }_{A}}\omega {{\partial }^{A}}\eta -{{{\rm e}}^{2u}}\omega ^{\prime} \eta ^{\prime} \right)}{\eta }.\end{eqnarray} \tag{ 10 }$$

The evolution equations for the metric q_AB (by equation (2); this is only one equation for the conformal factor u) and the second fundamental form ${{\chi }_{AB}}$ are given by

$$\begin{eqnarray}&&2\dot{u}={{\partial }_{A}}{{\beta }^{A}}+2{{\beta }^{A}}{{\partial }_{A}}u,\end{eqnarray} \tag{ 11 }$$

$$\begin{eqnarray}&&{{\dot{\chi }}_{AB}}={{\unicode{xa3}}_{\beta }}{{\chi }_{AB}}-{{F}_{AB}}-\alpha {{G}_{AB}}-2\alpha {{\chi }_{AC}}\chi _{B}^{C},\end{eqnarray} \tag{ 12 }$$

where $\unicode{xa3}$ denotes the Lie derivative and we have defined

$$\begin{eqnarray}&&{{F}_{AB}}={{\partial }_{A}}{{\partial }_{B}}\alpha -\frac{1}{2}{{\delta }_{AB}}\Delta \alpha -2{{\partial }_{\left( A \right.}}\alpha {{\partial }_{\left. B \right)}}u+{{\partial }_{C}}\alpha {{\partial }^{C}}u{{\delta }_{AB}},\end{eqnarray} \tag{ 13 }$$

and

$$\begin{eqnarray}&&{{G}_{AB}}={{\;}^{(3)}}{{R}_{AB}}-\frac{1}{2}{{\delta }_{AB}}{{\;}^{(3)}}{{R}_{CD}}{{\delta }^{CD}},\end{eqnarray} \tag{ 14 }$$

$$\begin{eqnarray}&&^{(3)}{{R}_{AB}}=\frac{1}{2{{\eta }^{2}}}({{\partial }_{A}}\eta {{\partial }_{B}}\eta +{{\partial }_{A}}\omega {{\partial }_{B}}\omega ).\end{eqnarray} \tag{ 15 }$$

Constraint equations: the momentum and Hamiltonian constraints are given by

$$\begin{eqnarray}&&{{\partial }^{B}}{{\chi }_{AB}}=-\frac{{{{\rm e}}^{2u}}}{2{{\eta }^{2}}}\left( \eta ^{\prime} {{\partial }_{A}}\eta +\omega ^{\prime} {{\partial }_{A}}\omega \right),\end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray}&&^{(3)}\Delta \sigma +\Delta q=-\frac{\varepsilon }{4\rho },\end{eqnarray} \tag{ 17 }$$

where we have defined the energy density ε by

$$\begin{eqnarray}&&\varepsilon =\left( \frac{{{{\rm e}}^{2u}}}{{{\eta }^{2}}}\left( \eta {{^{\prime} }^{2}}+\omega {{^{\prime} }^{2}} \right)+{{\left| \partial \sigma \right|}^{2}}+\frac{|\partial \omega {{|}^{2}}}{{{\eta }^{2}}}+2{{{\rm e}}^{-2u}}{{\chi }^{AB}}{{\chi }_{AB}} \right)\;\rho .\end{eqnarray} \tag{ 18 }$$

It is important to emphasize that ε is positive definite.

Gauge equations: the gauge equations for lapse and shift are given by

$$\begin{eqnarray}&&\Delta \alpha =\alpha \left( {{{\rm e}}^{-2u}}{{\chi }^{AB}}{{\chi }_{AB}}+{{{\rm e}}^{2u}}\bar{\mu } \right),\end{eqnarray} \tag{ 19 }$$

$$\begin{eqnarray}&&{{(\mathcal{L}\beta )}_{AB}}=2\alpha {{{\rm e}}^{-2u}}{{\chi }_{AB}},\end{eqnarray} \tag{ 20 }$$

where we have defined $\bar{\mu }$ by

$$\begin{eqnarray}&&\bar{\mu }=\frac{1}{2{{\eta }^{2}}}\left( \eta {{^{\prime} }^{2}}+\omega {{^{\prime} }^{2}} \right).\end{eqnarray} \tag{ 21 }$$

As we mentioned earlier, the most important property of this gauge is that the total ADM mass of the spacetime is given by the following integral on the half-plane $\mathbb{R}_{+}^{2}$ of the positive definite energy density ε

$$\begin{eqnarray}&&m=\frac{1}{16}{{\int }_{\mathbb{R}_{+}^{2}}}\varepsilon \;{\rm d}\rho {\rm d}z.\end{eqnarray} \tag{ 22 }$$

Moreover, this quantity is conserved along the evolution in this gauge (see [19]). We emphasize that the domain of integration in (22) is $\mathbb{R}_{+}^{2}$ even in the case of a black hole (see the discussion in [20]).

We have introduced two slight changes of notation with respect to [24]. First we have suppressed the hat symbol over tensors like ${{\hat{\chi }}^{AB}}$ introduced in [24] to distinguish between indices moved with the flat metric ${{\delta }_{AB}}$ and with the metric q_AB. In this article there is no danger of confusion since all the indices are moved with the flat metric ${{\delta }_{AB}}$. Second, we have defined the energy density ε in (18) with an extra factor ρ. This is convenient for the calculations presented in the next section, since the integral in the mass (22) then has the flat volume element in $\mathbb{R}_{+}^{2}$ (in [24], the ρ factor appears in the volume element). The only disadvantage of this notation is that in the right side of the Hamiltonian constraint (17), an extra ρ appears in the denominator.

Boundary conditions: at spacelike infinity we assume the following standard asymptotically flat fall-off condition in the limit $r\to \infty$

$$\begin{eqnarray}&&\sigma ,{{\beta }^{A}},{{\chi }_{AB}},\dot{\sigma },{{\dot{\beta }}^{A}},{{\dot{\chi }}_{AB}}={{o}_{1}}({{r}^{-1/2}}),\qquad \quad \bar{\alpha }-1={{o}_{1}}(1),\end{eqnarray} \tag{ 23 }$$

where we write $f={{o}_{j}}({{r}^{k}})$ if f satisfies ${{\partial }^{\alpha }}f=o({{r}^{k-|\alpha |}})$, for $|\alpha |\leqslant j$, where α is a multi-index and the spherical radius r is defined by $r=\sqrt{{{\rho }^{2}}+{{z}^{2}}}$. In the following, we will also make use of a similar notation for $f={{O}_{j}}({{r}^{k}})$.

At the axis, the functions must satisfy the following parity conditions:

$$\begin{eqnarray}&&\eta ,\ \omega ,\ \bar{\alpha },\ u,\ q,\ \sigma ,\ {{\chi }_{\rho \rho }},\ {{\beta }^{z}}\ {\rm are}\ {\rm even}\ {\rm functions}\ {\rm of}\ \rho \end{eqnarray} \tag{ 24 }$$

and

$$\begin{eqnarray}&&\alpha ,\ {{\chi }_{\rho z}},\ {{\beta }^{\rho }}\ {\rm are}\ {\rm odd}\ {\rm functions}\ {\rm of}\ \rho .\end{eqnarray} \tag{ 25 }$$

Note that odd functions vanish at the axis and the ρ derivative of even functions vanishes at the axis.

In the case of the extreme Kerr black hole, we have an extra asymptotic end, which in these coordinates is located at the origin. For that case, we will assume the following behavior in the limit $r\to 0$:

$$\begin{eqnarray}&&\sigma ,{{\beta }^{A}},{{\chi }_{AB}},\dot{\sigma },{{\dot{\beta }}^{A}},{{\dot{\chi }}_{AB}}={{o}_{1}}({{r}^{-1/2}}),\quad \bar{\alpha }-1={{o}_{1}}(1).\end{eqnarray} \tag{ 26 }$$

These conditions encompass the asymptotically cylindrical behavior typical of extreme black hole at this end (see the discussion in [20, 23]).

The behavior of the twist ω is more subtle because it contains the information of the angular momentum. It will be discussed in the next sections.

2.2. Linearization

Denote by $\psi$ any of the unknowns of the previous equations. Consider a one-parameter family of exact solutions $\psi (\lambda )$. To linearize the equations with respect to the family $\psi (\lambda )$ means to take a derivative with respect to λ to the equations and evaluate them at $\lambda =0$. We will use the following notation for the background and the first order linearization:

$$\begin{eqnarray}&&{{\psi }_{0}}={{\left. \psi (\lambda ) \right|}_{\lambda =0}},\qquad \quad {{\psi }_{1}}={{\left. \frac{{\rm d}\psi (\lambda )}{{\rm d}\lambda } \right|}_{\lambda =0}}.\end{eqnarray} \tag{ 27 }$$

We will assume that the background solution is stationary in this gauge, that is,

$$\begin{eqnarray}&&{{\dot{\psi }}_{0}}=0.\end{eqnarray} \tag{ 28 }$$

Moreover, we will also assume that the background shift and second fundamental form vanished

$$\begin{eqnarray}&&\beta _{0}^{A}=0,\qquad \quad {{\chi }_{0AB}}=0.\end{eqnarray} \tag{ 29 }$$

The condition (29) is satisfied by the Kerr solution for any choice of the mass and angular momentum parameters (see appendix A). This condition simplifies considerably the equations. In particular, from (28) and (29) we deduce

$$\begin{eqnarray}&&\psi {{^{\prime} }_{0}}=0.\end{eqnarray} \tag{ 30 }$$

The first important consequence of the background assumptions in (29) is that the first order expansion of the lapse is trivial. Namely, the right side of equation (19) is second order in λ; hence, we obtain

$$\begin{eqnarray}&&\Delta {{\alpha }_{0}}=0,\qquad \quad \Delta {{\alpha }_{1}}=0.\end{eqnarray} \tag{ 31 }$$

Since the boundary conditions for α are independent of λ, it follows that the first order perturbation α₁ satisfies the homogeneous boundary condition both at the axis and at infinity, and hence from equation (31) we obtain that

$$\begin{eqnarray}&&{{\alpha }_{1}}=0.\end{eqnarray} \tag{ 32 }$$

In contrast, the zero order lapse α₀ satisfies non-trivial boundary conditions. The specific value of α₀ will depend, of course, on the choice of background. Remarkably, for Minkowski and extreme Kerr we have ${{\alpha }_{0}}=\rho$, as we will see in the next sections. But for non-extreme Kerr, it has a different value (see appendix A). In this section we keep α₀ arbitrary in order to obtain general equations that can be used in future works for non-extreme black holes.

Using (32), (29), and (28) we find the following useful formulas:

$$\begin{eqnarray}&&\psi _{1}^{\prime }=\frac{1}{{{\alpha }_{0}}}\left( {{{\dot{\psi }}}_{1}}-\beta _{1}^{A}{{\partial }_{A}}{{\psi }_{0}} \right),\end{eqnarray} \tag{ 33 }$$

$$\begin{eqnarray}&&\psi _{1}^{\prime\prime }=\frac{1}{\alpha _{0}^{2}}\left( {{\ddot{\psi }}_{1}}-\dot{\beta }_{1}^{A}{{\partial }_{A}}{{\psi }_{0}} \right).\end{eqnarray} \tag{ 34 }$$

Also, as consequence of the definition in (4) we have the following relations between η and σ:

$$\begin{eqnarray}&&{{\eta }_{0}}={{\rho }^{2}}{{{\rm e}}^{{{\sigma }_{0}}}},\qquad \quad {{\eta }_{1}}={{\eta }_{0}}{{\sigma }_{1}}.\end{eqnarray} \tag{ 35 }$$

Using these assumptions, it is straightforward to obtain the linearization of the equations presented in section 2.1. The result is the following.

Evolution equations: the evolution equations for σ₁ and ω₁ are given by

$$\begin{eqnarray}&&-\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\alpha _{0}^{2}}\dot{p}{{+}^{(3)}}\Delta {{\sigma }_{1}}+\frac{{{\partial }_{A}}{{\sigma }_{1}}{{\partial }^{A}}\overline{{{\alpha }_{0}}}}{{{{\bar{\alpha }}}_{0}}}=\frac{2}{\eta _{0}^{2}}\left( {{\sigma }_{1}}|\partial {{\omega }_{0}}{{|}^{2}}-{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{\omega }_{0}} \right),\end{eqnarray} \tag{ 36 }$$

$$\begin{eqnarray}&&-\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\alpha _{0}^{2}}\dot{d}{{+}^{(3)}}\Delta {{\omega }_{1}}+\frac{{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{{\bar{\alpha }}}_{0}}}{{{{\bar{\alpha }}}_{0}}}=4\frac{{{\partial }_{\rho }}{{\omega }_{1}}}{\rho }+2{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{\sigma }_{0}}+2{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\sigma }_{1}},\end{eqnarray} \tag{ 37 }$$

where we have defined the following two useful auxiliary variables:

$$\begin{eqnarray}&&p=\mathop{{{\sigma }_{1}}}\limits^{.}-\beta _{1}^{A}{{\partial }_{A}}{{\sigma }_{0}}-2\frac{{{\beta }^{\rho }}}{\rho },\end{eqnarray} \tag{ 38 }$$

$$\begin{eqnarray}&&d=\mathop{{{\omega }_{1}}}\limits^{.}-\beta _{1}^{A}{{\partial }_{A}}{{\omega }_{0}}.\end{eqnarray} \tag{ 39 }$$

The evolution equation for the metric and second fundamental form are given by

$$\begin{eqnarray}&&2{{\dot{u}}_{1}}={{\partial }_{A}}\beta _{1}^{A}+2\beta _{1}^{A}{{\partial }_{A}}{{u}_{0}},\end{eqnarray} \tag{ 40 }$$

$$\begin{eqnarray}&&{{\dot{\chi }}_{1AB}}=-\left( {{F}_{1AB}}+{{\alpha }_{0}}{{G}_{1AB}} \right),\end{eqnarray} \tag{ 41 }$$

where

$$\begin{eqnarray}&&{{F}_{1AB}}=-2{{\partial }_{\left( A \right.}}{{\alpha }_{0}}{{\partial }_{\left. B \right)}}{{u}_{1}}+{{\delta }_{AB}}{{\partial }_{C}}{{\alpha }_{0}}{{\partial }^{C}}{{u}_{1}},\end{eqnarray} \tag{ 42 }$$

and

$$\begin{eqnarray}\begin{array}{rcl} {{G}_{1AB}} & = & \frac{1}{2\eta _{0}^{2}}\left( {{\partial }_{A}}{{\eta }_{1}}{{\partial }_{B}}{{\eta }_{0}}+{{\partial }_{A}}{{\eta }_{0}}{{\partial }_{B}}{{\eta }_{1}}+{{\partial }_{A}}{{\omega }_{1}}{{\partial }_{B}}{{\omega }_{0}}+{{\partial }_{A}}{{\omega }_{0}}{{\partial }_{B}}{{\omega }_{1}} \right) \\ {} & {} & -\frac{{{\sigma }_{1}}}{\eta _{0}^{2}}\left( {{\partial }_{A}}{{\eta }_{0}}{{\partial }_{B}}{{\eta }_{0}}+{{\partial }_{A}}{{\omega }_{0}}{{\partial }_{B}}{{\omega }_{0}} \right) \\ {} & {} & -\frac{{{\delta }_{AB}}}{2}\left[ \frac{1}{\eta _{0}^{2}}\left( {{\partial }_{C}}{{\eta }_{0}}{{\partial }^{C}}{{\eta }_{1}}+{{\partial }_{C}}{{\omega }_{0}}{{\partial }^{C}}{{\omega }_{1}} \right)-\frac{{{\sigma }_{1}}}{\eta _{0}^{2}}(|\partial {{\eta }_{0}}{{|}^{2}}+|\partial {{\omega }_{0}}{{|}^{2}}) \right]. \\ \end{array}\end{eqnarray} \tag{ 43 }$$

Constraint equations: the momentum constraint and Hamiltonian constraints are given by

$$\begin{eqnarray}&&{{\partial }^{B}}{{\chi }_{1AB}}=-\frac{{{{\rm e}}^{2{{u}_{0}}}}}{2{{\alpha }_{0}}}\left( p\left( {{\partial }_{A}}{{\sigma }_{0}}+2\frac{{{\partial }_{A}}\rho }{\rho } \right)+\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}}d \right),\end{eqnarray} \tag{ 44 }$$

$$\begin{eqnarray}&&^{(3)}\Delta {{\sigma }_{1}}+\Delta {{q}_{1}}=-\frac{{{\varepsilon }_{1}}}{4\rho },\end{eqnarray} \tag{ 45 }$$

where ${{\varepsilon }_{1}}$ is the first order term of the energy density in (18); that is

$$\begin{eqnarray}&&{{\varepsilon }_{1}}=\left( 2{{\partial }_{A}}{{\sigma }_{0}}{{\partial }^{A}}{{\sigma }_{1}}+\frac{2{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\omega }_{1}}}{\eta _{0}^{2}}-\frac{2{{\sigma }_{1}}|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}} \right)\;\rho .\end{eqnarray} \tag{ 46 }$$

Gauge equations: we have seen that the first order lapse is zero. For the shift we have

$$\begin{eqnarray}&&{{\left( \mathcal{L}{{\beta }_{1}} \right)}^{AB}}=2{{{\rm e}}^{-2{{u}_{0}}}}{{\alpha }_{0}}\chi _{1}^{AB}.\end{eqnarray} \tag{ 47 }$$

We have presented here the complete set of axially symmetric linear equations in the maximal-isothermal gauge. The conserved energy for this system of equations is calculated from the second variation of the energy density in (18) as follows. Assume that $\psi (\lambda )$ has the following form:

$$\begin{eqnarray}&&\psi (\lambda )={{\psi }_{0}}+\lambda {{\psi }_{1}}.\end{eqnarray} \tag{ 48 }$$

That is, we assume that the second order derivative with respect to λ of $\psi (\lambda )$ vanishes at $\lambda =0$. For this kind of linear perturbation, we define the second variation of ε as

$$\begin{eqnarray}&&{{\varepsilon }_{2}}={{\left. \frac{{{{\rm d}}^{2}}\varepsilon (\lambda )}{{\rm d}{{\lambda }^{2}}} \right|}_{\lambda =0}}.\end{eqnarray} \tag{ 49 }$$

Using (18) we obtain

$$\begin{eqnarray}\begin{array}{rcl} {{\varepsilon }_{2}} & = & \left( \frac{2{{{\rm e}}^{2{{u}_{0}}}}}{\alpha _{0}^{2}}\left( {{p}^{2}}+\frac{{{d}^{2}}}{\eta _{0}^{2}} \right)-8\frac{{{\sigma }_{1}}{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\omega }_{1}}}{\eta _{0}^{2}} \right. \\ {} & {} & +\left. 2{{\left| \partial {{\sigma }_{1}} \right|}^{2}}+4{{{\rm e}}^{-2{{u}_{0}}}}\chi _{1}^{AB}{{\chi }_{1AB}}+2\frac{|\partial {{\omega }_{1}}{{|}^{2}}}{\eta _{0}^{2}}+4\frac{|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}}\sigma _{1}^{2} \right)\rho . \\ \end{array}\end{eqnarray} \tag{ 50 }$$

Note that ${{\varepsilon }_{2}}$, in contrast with ε, is not positive definite.

For further reference we also write the zero order expression for the energy density

$$\begin{eqnarray}&&{{\varepsilon }_{0}}=\left( {{\left| \partial {{\sigma }_{0}} \right|}^{2}}+\frac{|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}} \right)\;\rho ,\end{eqnarray} \tag{ 51 }$$

and the masses associated with the different orders of the energy density

$$\begin{eqnarray}&&{{m}_{0}}=\frac{1}{16}{{\int }_{\mathbb{R}_{+}^{2}}}{{\varepsilon }_{0}}\;{\rm d}\rho {\rm d}z,\end{eqnarray} \tag{ 52 }$$

$$\begin{eqnarray}&&{{m}_{1}}=\frac{1}{16}{{\int }_{\mathbb{R}_{+}^{2}}}{{\varepsilon }_{1}}\;{\rm d}\rho {\rm d}z,\end{eqnarray} \tag{ 53 }$$

$$\begin{eqnarray}&&{{m}_{2}}=\frac{1}{16}{{\int }_{\mathbb{R}_{+}^{2}}}{{\varepsilon }_{2}}\;{\rm d}\rho {\rm d}z.\end{eqnarray} \tag{ 54 }$$

Recall that ${{\varepsilon }_{1}}$ has been calculated in (46).

We will prove that m₁ vanished and that m₂ is conserved and positive definite. Since we are interested in the study of linear stability, it is important for our present purpose (and also for future works on this subject) to prove these statements using only the linear equations, without referring to the original non-linear system. In the next sections we will perform these proofs. However, from the conceptual point of view and for further possible applications to the non-linear stability problem, it is important also to deduce these properties from the full equations. We discuss this point next.

Consider a general one-parameter family of exact solutions $\psi (\lambda )$ (i.e., we are not assuming the particular linear form in (48)). For this family we compute the exact mass $m(\lambda )$ given by equation (22). This quantity is conserved, that is

$$\begin{eqnarray}&&\frac{{\rm d}m(\lambda )}{{\rm d}t}=0.\end{eqnarray} \tag{ 55 }$$

This equation is valid for all λ. Taking derivatives with respect to λ of equation (55) and then evaluating them in $\lambda =0$, we obtain that

$$\begin{eqnarray}&&{{\left. \frac{{\rm d}}{{\rm d}t}m \right|}_{\lambda =0}}=0,\end{eqnarray} \tag{ 56 }$$

$$\begin{eqnarray}&&{{\left. \frac{{\rm d}}{{\rm d}t}\frac{{\rm d}m}{{\rm d}\lambda } \right|}_{\lambda =0}}=0,\end{eqnarray} \tag{ 57 }$$

$$\begin{eqnarray}&&{{\left. \frac{{\rm d}}{{\rm d}t}\frac{{{{\rm d}}^{2}}\;m}{{\rm d}{{\lambda }^{2}}} \right|}_{\lambda =0}}=0.\end{eqnarray} \tag{ 58 }$$

We can, of course, take more derivatives with respect to λ, but this will not provide any useful conserved quantity for the linear equations.

It is clear that equations (56) and (57) are precisely

$$\begin{eqnarray}&&\frac{{\rm d}{{m}_{0}}}{{\rm d}t}=0,\end{eqnarray} \tag{ 59 }$$

$$\begin{eqnarray}&&\frac{{\rm d}{{m}_{1}}}{{\rm d}t}=0,\end{eqnarray} \tag{ 60 }$$

where m₀ and m₁ are given by (52) and (53), respectively.

The first equation (59) asserts that the mass of the background metric is conserved. This is, of course, valid even when the background solution is not stationary. In our case, since the background metric is stationary, not only is m₀ conserved, but also, the integrand ε₀, given by equation (51), is time independent, and hence the conservation in (59) is trivial.

Since m₁ depends only on the background solution ${{\psi }_{0}}$ and the first order perturbation ${{\psi }_{1}}$ (recall that ${{\psi }_{0}}$ and ${{\psi }_{1}}$ are defined by (27) for a general family $\psi (\lambda )$), then equation (60) asserts that m₁ is a conserved quantity for the linear equations. That is, from the exact conservation law in (55), we have deduced the conservation of m₁ for the linear equations.

For a general background, m₁ will be non-zero. However, using the Hamiltonian formulation of general relativity, it is possible to show that the first variation of the ADM mass vanishes on stationary solutions (see [8] and references therein). In section 4 we explicitly perform this computation adapted to our settings.

For the third equation (58), the situation is different. This equation asserts that the quantity

$$\begin{eqnarray}&&{{\hat{m}}_{2}}={{\left. \frac{{{{\rm d}}^{2}}m}{{\rm d}{{\lambda }^{2}}} \right|}_{\lambda =0}},\end{eqnarray} \tag{ 61 }$$

is conserved

$$\begin{eqnarray}&&\frac{{\rm d}{{{\hat{m}}}_{2}}}{{\rm d}t}=0.\end{eqnarray} \tag{ 62 }$$

However, ${{\hat{m}}_{2}}$ depends on the background solution ${{\psi }_{0}}$, the linear perturbation ${{\psi }_{1}}$, but also on the second order perturbation

$$\begin{eqnarray}&&{{\psi }_{2}}=\frac{{{{\rm d}}^{2}}\psi (\lambda )}{{\rm d}{{\lambda }^{2}}}{{|}_{\lambda =0}}.\end{eqnarray} \tag{ 63 }$$

Then ${{\hat{m}}_{2}}$ is not a quantity that can be computed purely in terms of the background solution ${{\psi }_{0}}$ and the linear perturbation ${{\psi }_{1}}$, and hence it cannot be used for the linearized equations.

Note that the mass m₂ defined in (54) is computed only using first order perturbations (since we have assumed (48) to compute it). In principle, m₂ and ${{\hat{m}}_{2}}$ are different quantities. Hence, the conservation law

$$\begin{eqnarray}&&\frac{{\rm d}{{m}_{2}}}{{\rm d}t}=0,\end{eqnarray} \tag{ 64 }$$

cannot be deduced directly from (62). But, as we will prove next, it turns out that if the background is stationary and hence the first variation m₁ vanishes, then we have ${{\hat{m}}_{2}}={{m}_{2}}$.

Let us compute explicitly ${{\hat{m}}_{2}}$. We define

$$\begin{eqnarray}&&{{\hat{\varepsilon }}_{2}}={{\left. \frac{{{{\rm d}}^{2}}\varepsilon (\lambda )}{{\rm d}{{\lambda }^{2}}} \right|}_{\lambda =0}}.\end{eqnarray} \tag{ 65 }$$

We emphasize that in (65) we are not assuming (48), and hence this is different from (49). The difference between ${{\varepsilon }_{2}}$ and ${{\hat{\varepsilon }}_{2}}$ is given by

$$\begin{eqnarray}&&{{\hat{\varepsilon }}_{2}}-{{\varepsilon }_{2}}=\left( 2{{\partial }_{A}}{{\sigma }_{0}}{{\partial }^{A}}{{\sigma }_{2}}+\frac{2{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\omega }_{2}}}{\eta _{0}^{2}}-\frac{2{{\sigma }_{2}}|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}} \right)\;\rho .\end{eqnarray} \tag{ 66 }$$

In this calculation, we have assumed that the background is stationary in this gauge (namely, we have assumed (28) and (29)). The difference between ${{\varepsilon }_{2}}$ and ${{\hat{\varepsilon }}_{2}}$ involves, of course, the second order perturbation σ₂ and ω₂. However, the right-hand side of (66) is exactly the same expression as the first variation ${{\varepsilon }_{1}}$ if we replace σ₁ and ω₁ in ${{\varepsilon }_{1}}$ (given by (46)) with σ₂ and ω₂. Hence, if m₁ vanishes on stationary solutions, then ${{\hat{m}}_{2}}={{m}_{2}}$ (that is, the integral of the right-hand side of (66) vanishes). In fact, this result is general and well known in the calculus of variations with non-linear variations (see, for example, [30] p 267).

Finally, let us discuss the sign of the second variation m₂. On Minkowski, the positive mass theorem clearly implies that the second variation of the mass should be positive since flat space is a global minimum of the mass. In the extreme Kerr case, there is no obvious connection between the positivity of the mass and the second variation. However, it has been proved that the mass has a minimum at extreme Kerr under variations with fixed angular momentum [18, 20]. To prove the positivity of the second variation m₂ on extreme Kerr in section 4, we will use similar techniques as in those references. As we pointed out earlier, for our purpose, it is important to prove this in terms only of the linearized equations.

The natural first application of the linear equations obtained in section 2.2 is to study the linear stability of Minkowski in axial symmetry. The problem of linear stability of Minkowski, without any symmetry assumptions, was solved in [12], and the non-linear stability of Minkowski was finally proved in [13]. The purpose of this section is to provide an alternative proof of the linear stability of Minkowski in axial symmetry using the gauge presented in the previous section. This is given in theorem 3.1, which constitutes the main result of this section.

In comparison with the results in [12], theorem 3.1 has the obvious disadvantage that it only applies to axially symmetric perturbation. Moreover, in this theorem, only pointwise boundedness of the solution is proved and not precise decay rates as in [12]. However, the advantage of this result is that it makes use only of energy estimates that can be generalized to the black hole case, as we will see in section 4.

This system of linear equations was studied numerically in [24] and analytically in [25]. The main difficulty is that the system is formally singular at the axis where $\rho =0$. Theorem 3.1 generalizes those works by including the twist and, more importantly, by obtaining a pointwise estimate of the solution in terms of conserved energies. We explain this point in more detail later.

The Minkowski background satisfies the assumptions in (29). The value of the other background quantities are the following:

$$\begin{eqnarray}&&{{\omega }_{0}}={{q}_{0}}={{\sigma }_{0}}=0,\end{eqnarray} \tag{ 67 }$$

and

$$\begin{eqnarray}&&{{u}_{0}}={\rm ln} \rho ,\qquad \quad {{\eta }_{0}}={{\rho }^{2}},\qquad \quad {{\alpha }_{0}}=\rho .\end{eqnarray} \tag{ 68 }$$

Introducing the background quantities in (67)–(68) on the linearized equations obtained in section 2.2, we arrive at the following set of equations for the linear axially symmetric perturbations of Minkowski.

Evolution equations: the evolution equations for σ₁ and ω₁ are given by

$$\begin{eqnarray}&&-\dot{p}{{+}^{(3)}}\Delta {{\sigma }_{1}}=0,\end{eqnarray} \tag{ 69 }$$

$$\begin{eqnarray}&&-{{\ddot{\omega }}_{1}}{{+}^{(3)}}\Delta {{\omega }_{1}}=4\frac{{{\partial }_{\rho }}{{\omega }_{1}}}{\rho },\end{eqnarray} \tag{ 70 }$$

where we defined the auxiliary function p by

$$\begin{eqnarray}&&p={{\dot{\sigma }}_{1}}-\frac{2\beta _{1}^{\rho }}{\rho }.\end{eqnarray} \tag{ 71 }$$

The evolution equations for the metric and the extrinsic curvature are given by

$$\begin{eqnarray}&&2{{\dot{u}}_{1}}={{\partial }_{A}}\beta _{1}^{A}+2\frac{\beta _{1}^{\rho }}{\rho },\end{eqnarray} \tag{ 72 }$$

$$\begin{eqnarray}&&{{\dot{\chi }}_{1AB}}=2{{\partial }_{\left( A \right.}}{{q}_{1}}{{\partial }_{\left. B \right)}}\rho -{{\delta }_{AB}}{{\partial }_{\rho }}{{q}_{1}}.\end{eqnarray} \tag{ 73 }$$

Constraint equations: the momentum and the Hamiltonian constraints take the following form:

$$\begin{eqnarray}&&{{\partial }^{A}}{{\chi }_{1AB}}=-p{{\partial }_{B}}\rho ,\end{eqnarray} \tag{ 74 }$$

$$\begin{eqnarray}&&\Delta {{q}_{1}}{{+}^{(3)}}\Delta {{\sigma }_{1}}=0.\end{eqnarray} \tag{ 75 }$$

Gauge equations for lapse and shift: we have proved in section 2.2 that the first order lapse is zero. The equation for the shift is given by

$$\begin{eqnarray}&&{{(\mathcal{L}{{\beta }_{1}})}^{AB}}=\frac{2}{\rho }\chi _{1}^{AB}.\end{eqnarray} \tag{ 76 }$$

For the mass density, we have

$$\begin{eqnarray}&&{{\varepsilon }_{0}}={{\varepsilon }_{1}}=0,\end{eqnarray} \tag{ 77 }$$

and hence we have

$$\begin{eqnarray}&&{{m}_{0}}={{m}_{1}}=0.\end{eqnarray} \tag{ 78 }$$

The second order mass density is given by

$$\begin{eqnarray}&&{{\varepsilon }_{2}}=\left( 2{{p}^{2}}+2\frac{\dot{\omega }_{1}^{2}}{{{\rho }^{4}}}+2|\partial {{\sigma }_{1}}{{|}^{2}}+2\frac{|\partial {{\omega }_{1}}{{|}^{2}}}{{{\rho }^{4}}}+4\frac{\chi _{1}^{AB}{{\chi }_{1AB}}}{{{\rho }^{2}}} \right)\;\rho .\end{eqnarray} \tag{ 79 }$$

It is important to note that ${{\varepsilon }_{2}}$ in the particular case of the Minkowski background, is positive definite.

Before presenting the main result, let us first discuss two simple but important properties of this set of equations. The first one (which only holds for the Minkowski background) is that the equation for the twist ω₁ in (70) decouples completely from the other equations⁴ . Then, it is useful to split the density ε₂ in two terms

$$\begin{eqnarray}&&{{\varepsilon }_{2}}={{\varepsilon }_{\sigma }}+{{\varepsilon }_{\omega }},\end{eqnarray} \tag{ 80 }$$

where

$$\begin{eqnarray}&&{{\varepsilon }_{\sigma }}=\left( 2{{p}^{2}}+2|\partial {{\sigma }_{1}}{{|}^{2}}+4\frac{\chi _{1}^{AB}{{\chi }_{1AB}}}{{{\rho }^{2}}} \right)\rho ,\end{eqnarray} \tag{ 81 }$$

$$\begin{eqnarray}&&{{\varepsilon }_{\omega }}=2\frac{\dot{\omega }_{1}^{2}}{{{\rho }^{3}}}+2\frac{|\partial {{\omega }_{1}}{{|}^{2}}}{{{\rho }^{3}}},\end{eqnarray} \tag{ 82 }$$

and the corresponding masses

$$\begin{eqnarray}&&{{m}_{2}}={{m}_{\sigma }}+{{m}_{\omega }},\end{eqnarray} \tag{ 83 }$$

where

$$\begin{eqnarray}&&{{m}_{\sigma }}={{\int }_{\mathbb{R}_{+}^{2}}}{{\varepsilon }_{\sigma }}\;{\rm d}\rho {\rm d}z,\quad {{m}_{\omega }}={{\int }_{\mathbb{R}_{+}^{2}}}{{\varepsilon }_{\omega }}\;{\rm d}\rho {\rm d}z.\end{eqnarray} \tag{ 84 }$$

Note that all the densities are positive definite.

Equation (70) is equivalent to the following homogeneous wave equation:

$$\begin{eqnarray}&&-{{\ddot{{\bar{\omega }}}}_{1}}{{+}^{(7)}}\Delta {{\bar{\omega }}_{1}}=0,\end{eqnarray} \tag{ 85 }$$

where $^{(7)}\Delta$ is the Laplacian in seven dimensions acting on axially symmetric functions⁵ , namely

$$\begin{eqnarray}&&^{(7)}\Delta {{\bar{\omega }}_{1}}=\Delta {{\bar{\omega }}_{1}}+5\frac{{{\partial }_{\rho }}{{{\bar{\omega }}}_{1}}}{\rho },\end{eqnarray} \tag{ 86 }$$

and we have defined

$$\begin{eqnarray}&&{{\bar{\omega }}_{1}}=\frac{{{\omega }_{1}}}{{{\rho }^{4}}}.\end{eqnarray} \tag{ 87 }$$

That is, the dynamic of the twist potential is determined by a wave equation, and hence it is clear how to obtain decay estimates for the solution. In contrast, the equations for σ₁ are coupled and non-standard due to the formal singular behavior at the axis (see the discussion in [24] and [25]). The wave equation in (85) has associated the canonical energy density

$$\begin{eqnarray}&&{{\epsilon }_{{\bar{\omega }}}}=2\left( \mathop{{\bar{\omega }}}\limits^{.}_{1}^{2}+|\partial {{{\bar{\omega }}}_{1}}{{|}^{2}} \right){{\rho }^{5}},\end{eqnarray} \tag{ 88 }$$

and corresponding energy

$$\begin{eqnarray}&&{{m}_{{\bar{\omega }}}}={{\int }_{{{\mathbb{R}}^{3}}}}{{\epsilon }_{{\bar{\omega }}}}\;{\rm d}\rho {\rm d}z.\end{eqnarray} \tag{ 89 }$$

The factor ρ⁵ in (88) comes from the expression of the volume element in seven dimensions in terms of the cylindrical coordinates ${\rm d}{{x}^{7}}={{\rho }^{5}}{\rm d}\rho {\rm d}z$. The two densities ${{\varepsilon }_{{\bar{\omega }}}}$ and ${{\varepsilon }_{\omega }}$ are related by a boundary term

$$\begin{eqnarray}&&{{\epsilon }_{{\bar{\omega }}}}-{{\varepsilon }_{\omega }}=-4{{\partial }_{\rho }}\left( \frac{\omega _{1}^{2}}{{{\rho }^{4}}} \right),\end{eqnarray} \tag{ 90 }$$

and hence ${{m}_{{\bar{\omega }}}}={{m}_{\omega }}$ provided ω₁ satisfies appropriate boundary conditions. Note that equation (85) suggests that ${{\bar{\omega }}_{1}}$ and not ω₁ is the most convenient variable to impose the boundary conditions.

The second property (which will also be satisfied for the Kerr background and in general for any stationary background) is the following. The coefficients of the equations do not depend on time; hence, if we take a time derivative to all equations, we get a new set of equations for the time derivatives of the unknowns that are formally identical to the original ones. That is, the variables ${{\sigma }_{1}},{{\omega }_{1}},{{u}_{1}},{{\beta }_{1}},{{\chi }_{1}}$ satisfy the same equations as the time derivatives ${{\dot{\sigma }}_{1}},{{\dot{\omega }}_{1}},{{\dot{u}}_{1}},{{\dot{\beta }}_{1}},{{\dot{\chi }}_{1}}$. And the same is, of course, true for any number of time derivatives. In particular, if m is a conserved quantity, then we automatically get an infinite number of conserved quantities that have the same form as m but in terms of the nth time derivatives of ${{\sigma }_{1}},{{\omega }_{1}},{{u}_{1}},{{\beta }_{1}},{{\chi }_{1}}$. For example, let us consider the mass ${{m}_{\sigma }}$ defined by (81) and (84). It depends on the functions p, σ₁ and χ₁. To emphasize this dependence we use the notation ${{m}_{\sigma }}[p,{{\sigma }_{1}},{{\chi }_{1}}]$. Then we define ${{m}_{\sigma }}[\dot{p},{{\dot{\sigma }}_{1}},{{\dot{\chi }}_{1}}]$ as

$$\begin{eqnarray}&&{{m}_{\sigma }}\left[ \dot{p},{{{\dot{\sigma }}}_{1}},{{{\dot{\chi }}}_{1}} \right]={{\int }_{\mathbb{R}_{+}^{2}}}\left( 2{{{\dot{p}}}^{2}}+2|\partial {{{\dot{\sigma }}}_{1}}{{|}^{2}}+4\frac{\dot{\chi }_{1}^{AB}{{{\dot{\chi }}}_{1AB}}}{{{\rho }^{2}}} \right)\;\rho {\rm d}\rho {\rm d}z.\end{eqnarray} \tag{ 91 }$$

If ${{m}_{\sigma }}[p,{{\sigma }_{1}},{{\chi }_{1}}]$ is conserved along the evolution, then ${{m}_{\sigma }}[\dot{p},{{\dot{\sigma }}_{1}},{{\dot{\chi }}_{1}}]$ is also conserved. The same applies for ${{m}_{\omega }}[{{\omega }_{1}}]$ and ${{m}_{{\bar{\omega }}}}[{{\bar{\omega }}_{1}}]$; for example, we have

$$\begin{eqnarray}&&{{m}_{{\bar{\omega }}}}\left[ {{\mathop{{\bar{\omega }}}\limits^{.}}_{1}} \right]={{\int }_{\mathbb{R}_{+}^{2}}}\left( \ddot{{\bar{\omega }}}_{1}^{2}+|\partial {{\mathop{{\bar{\omega }}}\limits^{.}}_{1}}{{|}^{2}} \right){{\rho }^{5}}{\rm d}\rho {\rm d}z.\end{eqnarray} \tag{ 92 }$$

We will also make use of the higher order masses ${{m}_{{\bar{\omega }}}}[{{\ddot{{\bar{\omega }}}}_{1}}]$ and ${{m}_{{\bar{\omega }}}}[\mathop{\mathop{\omega }\limits^{1\ldots \bar{\ }}}\limits_{.}^{}]$.

Theorem 3.1. Consider a smooth solution of the linearized equations presented earlier that satisfies the fall-off conditions at infinity (23) and the regularity conditions at the axis (24), (25). Assume also that

$$\begin{eqnarray}&&{{\mathop{{\bar{\omega }}}\limits^{.}}_{1}},{{\bar{\omega }}_{1}}={{O}_{1}}(1),\end{eqnarray} \tag{ 93 }$$

at the axis and

$$\begin{eqnarray}&&{{\mathop{{\bar{\omega }}}\limits^{.}}_{1}},{{\bar{\omega }}_{1}}={{o}_{1}}({{r}^{-5/2}}),\end{eqnarray} \tag{ 94 }$$

at infinity, where we have defined

$$\begin{eqnarray}&&{{\bar{\omega }}_{1}}=\frac{{{\omega }_{1}}}{{{\rho }^{4}}}.\end{eqnarray} \tag{ 95 }$$

Then, we have:

• (i)  
  The masses ${{m}_{\sigma }}$, ${{m}_{\omega }}$ and ${{m}_{{\bar{\omega }}}}$ defined by (84) and (89) are conserved along the evolution and ${{m}_{\omega }}={{m}_{{\bar{\omega }}}}$. And hence, all higher order masses are also conserved.
• (ii)  
  The solution ${{\sigma }_{1}},{{\omega }_{1}}$ satisfies the following (time independent) bounds:

  $$\begin{eqnarray}&&C|{{\sigma }_{1}}|\leqslant {{m}_{\sigma }}[p,{{\sigma }_{1}},{{\chi }_{1}}]+{{m}_{\sigma }}[\dot{p},{{\dot{\sigma }}_{1}},{{\dot{\chi }}_{1}}],\end{eqnarray} \tag{ 96 }$$

  $$\begin{eqnarray}&&C\frac{|{{\omega }_{1}}|}{{{\rho }^{4}}}\leqslant {{m}_{\omega }}\left[ {{\ddot{\omega }}_{1}} \right]+{{m}_{\omega }}[{{}_{1}}],\end{eqnarray} \tag{ 97 }$$

  where $C\gt 0$ is a numerical constant.

The value of ω at the axis determines the angular momentum (see, for example, [20]). Hence, the physical interpretation of the boundary conditions in (93) is that the perturbations do not change the angular momentum of the background (which is zero in the case of Minkowski).

The conservation of ${{m}_{\sigma }}$ in point (i) was proved in [24] . For completeness, we review this proof and also perform it using different variables, which are the appropriate ones for the extreme Kerr black hole case treated in the next section.

We have already shown that the equation for ω₁ is decoupled and it can be converted into a standard wave equation in higher dimensions. Hence, the dynamics of ω₁ is well known. In particular, one has the classical pointwise estimates for solutions of the wave equation in seven dimensions $|{{\bar{\omega }}_{1}}|\leqslant {{t}^{-3}}C$, where the constant C depends only on the initial data (see, for example, [42]). We present the weaker estimate in (97) because it can be proved using only the conserved energies and is likely to be useful in the more complex case of the Kerr black hole, where the pure wave equation estimates are not available.

The most important part of theorem 3.1 is the estimate in (96). In [25], the existence of a solution of this set of equations was proved using an explicit (but rather complicated) representation in terms of integral transforms. In contrast, the a priori estimate in (96) is proved in terms of only the conserved masses in a remarkably simple way. This estimate is expected to be useful in the non-linear regime.

Proof. (i) Since the equations are decoupled, we can treat the conservation for ${{m}_{\sigma }}$ and ${{m}_{\omega }}$ separately. We begin with ${{m}_{\sigma }}$. Taking the time derivative of ${{\varepsilon }_{\sigma }}$ we obtain

$$\begin{eqnarray}&&{{\dot{\varepsilon }}_{\sigma }}=4\rho p\dot{p}+4\rho {{\partial }_{A}}{{\sigma }_{1}}{{\partial }^{A}}{{\dot{\sigma }}_{1}}+8\frac{\chi _{1}^{AB}{{{\dot{\chi }}}_{1AB}}}{\rho }.\end{eqnarray} \tag{ 98 }$$

The strategy proves (using the linearized equations) that the right side of (98) is a total divergence, and hence it integrates to zero under appropriate boundary conditions. We calculate each terms individually.

For the first term, we just use the definition of p given in equation (71) to obtain

$$\begin{eqnarray}&&4\rho p\dot{p}=4\rho {{\dot{\sigma }}_{1}}\dot{p}-8\beta _{1}^{\rho }\dot{p}.\end{eqnarray} \tag{ 99 }$$

For the second term, we obtain

$$\begin{eqnarray}&&4\rho {{\partial }_{A}}{{\sigma }_{1}}{{\partial }^{A}}{{\dot{\sigma }}_{1}}=4{{\partial }^{A}}\left( \rho {{{\dot{\sigma }}}_{1}}{{\partial }_{A}}{{\sigma }_{1}} \right)-4{{\dot{\sigma }}_{1}}{{\partial }^{A}}\left( \rho {{\partial }_{A}}{{\sigma }_{1}} \right)\end{eqnarray} \tag{ 100 }$$

$$\begin{eqnarray}&&=\;4{{\partial }^{A}}\left( \rho {{{\dot{\sigma }}}_{1}}{{\partial }_{A}}{{\sigma }_{1}} \right)-4\rho {{\dot{\sigma }}_{1}}{{\;}^{(3)}}\Delta {{\sigma }_{1}}\end{eqnarray} \tag{ 101 }$$

$$\begin{eqnarray}&&=\;4{{\partial }^{A}}\left( \rho {{{\dot{\sigma }}}_{1}}{{\partial }_{A}}{{\sigma }_{1}} \right)-4\rho {{\dot{\sigma }}_{1}}\dot{p},\end{eqnarray} \tag{ 102 }$$

where in line (101) we have used the definition of the operator $^{(3)}\Delta$ given in equation (6) and in line (102) we have used equation (69).

Finally, for the third term, we have

$$\begin{eqnarray}&&8\frac{\chi _{1}^{AB}{{{\dot{\chi }}}_{1AB}}}{\rho }=4{{(\mathcal{L}{{\beta }_{1}})}^{AB}}{{\dot{\chi }}_{1AB}}\end{eqnarray} \tag{ 103 }$$

$$\begin{eqnarray}&&=\;8{{\partial }^{A}}\beta _{1}^{B}{{\dot{\chi }}_{1AB}}\end{eqnarray} \tag{ 104 }$$

$$\begin{eqnarray}&&=\;8{{\partial }^{A}}\left( \beta _{1}^{B}{{{\dot{\chi }}}_{1AB}} \right)-8\beta _{1}^{B}{{\partial }^{A}}{{\dot{\chi }}_{1AB}}\end{eqnarray} \tag{ 105 }$$

$$\begin{eqnarray}&&=\;8{{\partial }^{A}}\left( \beta _{1}^{B}{{{\dot{\chi }}}_{1AB}} \right)+8\dot{p}\beta _{1}^{\rho },\end{eqnarray} \tag{ 106 }$$

where in line (103) we have used the gauge equation in (76); in line (104), the fact that ${{\chi }_{1AB}}$ is trace-free; and in line (105), we have used the time derivative of equation (74).

Summing these results, we see that only the total divergence terms remain. That is

$$\begin{eqnarray}&&{{\dot{\varepsilon }}_{\sigma }}={{\partial }_{A}}{{t}^{A}},\end{eqnarray} \tag{ 107 }$$

where

$$\begin{eqnarray}&&{{t}_{A}}=4\rho {{\dot{\sigma }}_{1}}{{\partial }_{A}}{{\sigma }_{1}}+8\beta _{1}^{B}{{\dot{\chi }}_{1AB}}.\end{eqnarray} \tag{ 108 }$$

We integrate (107) in the half-disk D_L of radius L in $\mathbb{R}_{+}^{2}$, where C_L denotes the semi-circle of radius L (see figure 1). Using the divergence theorem in two dimensions, we obtain

$$\begin{eqnarray}&&{{\int }_{{{D}_{L}}}}{{\dot{\varepsilon }}_{\sigma }}\;{\rm d}\rho {\rm d}z={{\int }_{{{D}_{L}}}}{{\partial }_{A}}{{t}^{A}}\;{\rm d}\rho {\rm d}z\end{eqnarray} \tag{ 109 }$$

$$\begin{eqnarray}&&\qquad \qquad \quad {\mkern 1mu} ={{\int }_{\partial {{D}_{L}}}}{{t}^{A}}{{n}_{A}}\;{\rm d}s\end{eqnarray} \tag{ 110 }$$

$$\begin{eqnarray}&&\qquad \qquad \quad {\mkern 1mu} =-\int _{-L}^{L}{{t}^{\rho }}{{|}_{\rho =0}}\;{\rm d}z+{{\int }_{{{C}_{L}}}}{{t}^{A}}{{n}_{A}}\;{\rm d}s.\end{eqnarray} \tag{ 111 }$$

where n^A is the outwards unit normal vector and ${\rm d}s$ is the line element of C_L.

The integral of the first term in line (111) is given by

$$\begin{eqnarray}&&{{t}_{\rho }}=4\rho {{\dot{\sigma }}_{1}}{{\partial }_{\rho }}{{\sigma }_{1}}+8\beta _{1}^{\rho }{{\dot{\chi }}_{1\rho \rho }}+8\beta _{1}^{z}{{\dot{\chi }}_{1\rho z}}.\end{eqnarray} \tag{ 112 }$$

The first term clearly vanished at the axis $\rho =0$. The second and third term also vanish at the axis, since the regularity conditions in (25) imply that $\beta _{1}^{\rho }$ and ${{\chi }_{1\rho z}}$ are zero at the axis. Hence, we obtain

$$\begin{eqnarray}&&{{\int }_{{{D}_{L}}}}{{\dot{\varepsilon }}_{\sigma }}\;{\rm d}\rho {\rm d}z={{\int }_{{{C}_{L}}}}{{t}^{A}}{{n}_{A}}\;{\rm d}s.\end{eqnarray} \tag{ 113 }$$

Taking the limit $L\to \infty$ and using the fall-off conditions in (23) we obtain that the integral vanished, and hence ${{\dot{m}}_{\sigma }}=0$ (recall that on C_L we have ${\rm d}s={\rm d}r\theta$ where ${\rm tan} \theta =z/\rho$).

The conservation of ${{m}_{\omega }}$ is similar. We take the time derivative of the mass density ${{\varepsilon }_{\omega }}$

$$\begin{eqnarray}&&{{\dot{\varepsilon }}_{\omega }}=4\frac{{{{\dot{\omega }}}_{1}}{{\ddot{\omega }}_{1}}}{{{\rho }^{3}}}+4\frac{{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{{\dot{\omega }}}_{1}}}{{{\rho }^{3}}}.\end{eqnarray} \tag{ 114 }$$

For the first term, we have

$$\begin{eqnarray}&&4\frac{{{{\dot{\omega }}}_{1}}{{\ddot{\omega }}_{1}}}{{{\rho }^{3}}}=4\frac{{{{\dot{\omega }}}_{1}}}{{{\rho }^{3}}}\left( {{\partial }^{A}}{{\partial }_{A}}{{\omega }_{1}}-\frac{3{{\partial }_{\rho }}{{\omega }_{1}}}{\rho } \right),\end{eqnarray} \tag{ 115 }$$

where we have used equation (70).

For the second term, we have

$$\begin{eqnarray}&&4\frac{{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{{\dot{\omega }}}_{1}}}{{{\rho }^{3}}}=4{{\partial }^{A}}\left( \frac{{{{\dot{\omega }}}_{1}}{{\partial }_{A}}{{\omega }_{1}}}{{{\rho }^{3}}} \right)-4\frac{{{{\dot{\omega }}}_{1}}}{{{\rho }^{3}}}\left( {{\partial }^{A}}{{\partial }_{A}}{{\omega }_{1}}-\frac{3{{\partial }_{\rho }}{{\omega }_{1}}}{\rho } \right).\end{eqnarray} \tag{ 116 }$$

Hence, we obtain

$$\begin{eqnarray}&&{{\dot{\varepsilon }}_{\omega }}={{\partial }_{A}}{{t}^{A}},\end{eqnarray} \tag{ 117 }$$

with

$$\begin{eqnarray}&&{{t}_{A}}=4\frac{{{{\dot{\omega }}}_{1}}{{\partial }_{A}}{{\omega }_{1}}}{{{\rho }^{3}}}.\end{eqnarray} \tag{ 118 }$$

Integrating in the same domain as earlier and using the behavior at the axis in (93) and the fall-off conditions in (94) at infinity, we obtain that ${{\dot{m}}_{\omega }}=0$. Finally, the equality ${{m}_{\omega }}={{m}_{{\bar{\omega }}}}$ is deduced from (90) and the assumption from (93).

(ii) To prove the estimate (96) note that we have the following bounds

$$\begin{eqnarray}&&{{m}_{\sigma }}\left[ p,{{\sigma }_{1}},{{\chi }_{1}} \right]\geqslant {{\int }_{\mathbb{R}_{+}^{2}}}{{\left| \partial {{\sigma }_{1}} \right|}^{2}}\rho \;{\rm d}\rho {\rm d}z,\end{eqnarray} \tag{ 119 }$$

$$\begin{eqnarray}&&{{m}_{\sigma }}\left[ \dot{p},{{{\dot{\sigma }}}_{1}},{{{\dot{\chi }}}_{1}} \right]\geqslant 2{{\int }_{\mathbb{R}_{+}^{2}}}{{\dot{p}}^{2}}\rho \;{\rm d}\rho {\rm d}z=2{{\int }_{\mathbb{R}_{+}^{2}}}{{\left( ^{(3)}\Delta {{\sigma }_{1}} \right)}^{2}}\rho \;{\rm d}\rho {\rm d}z,\end{eqnarray} \tag{ 120 }$$

where in the last equality of line (120) we have used equation (69). The right side of (120) can be written in the following form:

$$\begin{eqnarray}&&{{\int }_{\mathbb{R}_{+}^{2}}}{{\left( ^{(3)}\Delta {{\sigma }_{1}} \right)}^{2}}\rho \;{\rm d}\rho {\rm d}z={{\int }_{{{\mathbb{R}}^{3}}}}{{\left( ^{(3)}\Delta {{\sigma }_{1}} \right)}^{2}}\;{\rm d}{{x}^{3}}\end{eqnarray} \tag{ 121 }$$

$$\begin{eqnarray}&&={{\int }_{{{\mathbb{R}}^{3}}}}|{{\partial }^{2}}{{\sigma }_{1}}{{|}^{2}}\;{\rm d}{{x}^{3}}.\end{eqnarray} \tag{ 122 }$$

where in the right side of line (121) we changed from cylindrical coordinates $(\rho ,z)$ to Cartesian coordinates $(x,y,z)$ in ${{\mathbb{R}}^{3}}$, with $x=\rho {\rm cos} \phi$, $y=\rho {\rm sin} \phi$. For axially symmetric functions (i.e., functions in ${{\mathbb{R}}^{3}}$ that do not depend on ϕ) we have ${\rm d}{{x}^{3}}=\rho \;{\rm d}\rho {\rm d}z$. In Cartesian coordinates, the Laplacian $^{(3)}\Delta$ is given by

$$\begin{eqnarray}&&^{(3)}\Delta {{\sigma }_{1}}=\partial _{x}^{2}{{\sigma }_{1}}+\partial _{y}^{2}{{\sigma }_{1}}+\partial _{z}^{2}{{\sigma }_{1}}.\end{eqnarray} \tag{ 123 }$$

And in line (122) we have integrated by parts; due to the fall-off assumptions on σ₁, the boundary term vanishes. In this equation $|{{\partial }^{2}}{{\sigma }_{1}}{{|}^{2}}$ denotes the sum of the squares of all second derivatives in terms of the Cartesian coordinates in ${{\mathbb{R}}^{3}}$, that is

$$\begin{eqnarray}&&{{\left| {{\partial }^{2}}{{\sigma }_{1}} \right|}^{2}}={{\left( \partial _{x}^{2}{{\sigma }_{1}} \right)}^{2}}+{{\left( \partial _{y}^{2}{{\sigma }_{1}} \right)}^{2}}+{{\left( \partial _{z}^{2}{{\sigma }_{1}} \right)}^{2}}+{{\left( {{\partial }_{x}}{{\partial }_{y}}{{\sigma }_{1}} \right)}^{2}}+{{\left( {{\partial }_{x}}{{\partial }_{z}}{{\sigma }_{1}} \right)}^{2}}+{{\left( {{\partial }_{y}}{{\partial }_{z}}{{\sigma }_{1}} \right)}^{2}}.\end{eqnarray} \tag{ 124 }$$

From (122), (120), and (119) we obtain the following crucial estimate:

$$\begin{eqnarray}&&{{m}_{\sigma }}\left[ p,{{\sigma }_{1}},{{\chi }_{1}} \right]+{{m}_{\sigma }}\left[ \dot{p},{{{\dot{\sigma }}}_{1}},{{{\dot{\chi }}}_{1}} \right]\geqslant {{\int }_{{{\mathbb{R}}^{3}}}}\left( |{{\partial }^{2}}{{\sigma }_{1}}{{|}^{2}}+|\partial {{\sigma }_{1}}{{|}^{2}} \right)\;{\rm d}{{x}^{3}}.\end{eqnarray} \tag{ 125 }$$

Note that on the right side of (125) there are no terms with $\sigma _{1}^{2}$, and hence we cannot directly use the standard Sobolev estimate to pointwise control the solution σ₁. However, using the estimate given by lemma B.1 with n = 3 and k = 2 we obtain the desired result in (96).

To obtain the estimate in (97) for ${{\bar{\omega }}_{1}}$, we proceed in a similar manner. From the definition of ${{m}_{{\bar{\omega }}}}$ we obtain

$$\begin{eqnarray}&&{{m}_{{\bar{\omega }}}}\geqslant {{\int }_{\mathbb{R}_{+}^{2}}}{{\left| \partial {{{\bar{\omega }}}_{1}} \right|}^{2}}{{\rho }^{5}}\;{\rm d}\rho {\rm d}z={{\int }_{{{\mathbb{R}}^{7}}}}{{\left| \partial {{{\bar{\omega }}}_{1}} \right|}^{2}}\;{\rm d}{{x}^{7}},\end{eqnarray} \tag{ 126 }$$

where we have used ${\rm d}{{x}^{7}}={{\rho }^{5}}{\rm d}\rho {\rm d}z$. For the higher order masses, we have

$$\begin{eqnarray}&&{{m}_{{\bar{\omega }}}}\left[ {{\mathop{{\bar{\omega }}}\limits^{.}}_{1}} \right]\geqslant {{\int }_{\mathbb{R}_{+}^{2}}}\ddot{{\bar{\omega }}}_{1}^{2}{{\rho }^{5}}\;{\rm d}\rho {\rm d}z\end{eqnarray} \tag{ 127 }$$

$$\begin{eqnarray}&&={{\int }_{{{\mathbb{R}}^{7}}}}{{\left( ^{(7)}\Delta {{{\bar{\omega }}}_{1}} \right)}^{2}}\;{\rm d}{{x}^{7}}\end{eqnarray} \tag{ 128 }$$

$$\begin{eqnarray}&&={{\int }_{{{\mathbb{R}}^{7}}}}|{{\partial }^{2}}{{\bar{\omega }}_{1}}{{|}^{2}}\;{\rm d}{{x}^{7}},\end{eqnarray} \tag{ 129 }$$

where in line (128) we have used the wave equation (85), and in line (129) we have integrated by parts and used that ${{\bar{\omega }}_{1}}$ decay at infinity. In a similar way, we obtain that energies with n-time derivatives control $n+1$ spatial derivatives; in particular

$$\begin{eqnarray}&&{{m}_{{\bar{\omega }}}}\left[ {{\ddot{{\bar{\omega }}}}_{1}} \right]\geqslant {{\int }_{{{\mathbb{R}}^{7}}}}{{\left| {{\partial }^{3}}{{{\bar{\omega }}}_{1}} \right|}^{2}}\;{\rm d}{{x}^{7}},\end{eqnarray} \tag{ 130 }$$

$$\begin{eqnarray}&&{{m}_{{\bar{\omega }}}}\left[ {{\mathop{{\bar{\omega }}}\limits^{.}}_{1}} \right]\geqslant {{\int }_{{{\mathbb{R}}^{7}}}}{{\left| {{\partial }^{4}}{{{\bar{\omega }}}_{1}} \right|}^{2}}\;{\rm d}{{x}^{7}}.\end{eqnarray} \tag{ 131 }$$

Using the bound (130), (131), and lemma B.1 with n = 7 and k = 4, the estimate in (97) follows. □

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We finally remark that in the proof of the conservation of m₂ we have used only the evolution equations for σ₁ and ω₁, the time derivative of the momentum constraint, and the gauge equation for the shift.

In this section we study the linearized equation obtained in section 2.2 for the case of the extreme Kerr background. The main difference with respect to the previous case of Minkowski is that the background quantities ${{q}_{0}},{{\sigma }_{0}},{{\omega }_{0}}$ are not zero. However, we still have (see appendix A)

$$\begin{eqnarray}&&{{\alpha }_{0}}=\rho .\end{eqnarray} \tag{ 132 }$$

This is the main remarkable simplification of the extreme Kerr case compared with the non-extreme Kerr black hole.

For the explicit form of $({{q}_{0}},{{\sigma }_{0}},{{\omega }_{0}})$, see appendix A. These functions depend on one parameter: the mass m₀ of the black hole. This mass is given by (52). The only properties of these functions that we will use are the following. They satisfy the stationary equations

$$\begin{eqnarray}&&^{(3)}\Delta {{\sigma }_{0}}=\frac{|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}},\end{eqnarray} \tag{ 133 }$$

$$\begin{eqnarray}&&{{\partial }^{A}}\left( \frac{\rho {{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)=0.\end{eqnarray} \tag{ 134 }$$

They satisfy the fall-off conditions in (23) and (26). They satisfy the following inequality in $\mathbb{R}_{+}^{2}$ (i.e., including both the origin and infinity) (see [18])

$$\begin{eqnarray}&&\frac{|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}}\leqslant \frac{C}{{{r}^{2}}},\qquad \quad {{\left| \partial {{\sigma }_{0}} \right|}^{2}}\leqslant \frac{C}{{{r}^{2}}},\end{eqnarray} \tag{ 135 }$$

where C is a constant that depends only on m₀. Finally, near the axis, we have

$$\begin{eqnarray}&&\frac{{{\partial }_{\rho }}{{\omega }_{0}}}{{{\eta }_{0}}}=O\left( \rho \right).\end{eqnarray} \tag{ 136 }$$

The complete set of linearized equations, in axial symmetry, for the extreme Kerr black hole is the following.

Evolution equations: the evolution equations for σ₁ and ω₁ are given by

$$\begin{eqnarray}&&-\frac{{{e}^{2{{u}_{0}}}}}{{{\rho }^{2}}}\dot{p}{{+}^{(3)}}\Delta {{\sigma }_{1}}=\frac{2}{\eta _{0}^{2}}\left( {{\sigma }_{1}}|\partial {{\omega }_{0}}{{|}^{2}}-{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{\omega }_{0}} \right),\end{eqnarray} \tag{ 137 }$$

$$\begin{eqnarray}&&-\frac{{{e}^{2{{u}_{0}}}}}{{{\rho }^{2}}}\dot{d}{{+}^{(3)}}\Delta {{\omega }_{1}}=4\frac{{{\partial }_{\rho }}{{\omega }_{1}}}{\rho }+2{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{\sigma }_{0}}+2{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\sigma }_{1}},\end{eqnarray} \tag{ 138 }$$

with

$$\begin{eqnarray}&&p={{\dot{\sigma }}_{1}}-2\frac{\beta _{1}^{\rho }}{\rho }-\beta _{1}^{A}{{\partial }_{A}}{{\sigma }_{0}},\end{eqnarray} \tag{ 139 }$$

$$\begin{eqnarray}&&d={{\dot{\omega }}_{1}}-\beta _{1}^{A}{{\partial }_{A}}{{\omega }_{0}}.\end{eqnarray} \tag{ 140 }$$

The evolution equations for the metric and the second fundamental are obtained, replacing (132) in equations (40) and (41). No relevant simplification occurs in these equations compared with the general expressions (40) and (41), and hence we do not write them again in this section. Also, we will not make use of these equations in the proof of theorem 4.1.

Constraint equations: the momentum constraint and Hamiltonian constraint are given by

$$\begin{eqnarray}&&{{\partial }^{B}}{{\chi }_{1AB}}=-\frac{{{{\rm e}}^{2{{u}_{0}}}}}{2\rho }\left( p\left( 2\frac{{{\partial }_{A}}\rho }{\rho }+{{\partial }_{A}}{{\sigma }_{0}} \right)+\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}}d \right),\end{eqnarray} \tag{ 141 }$$

$$\begin{eqnarray}&&^{(3)}\Delta {{\sigma }_{1}}+\Delta {{q}_{1}}=-\frac{{{\varepsilon }_{1}}}{4\rho },\end{eqnarray} \tag{ 142 }$$

where ${{\varepsilon }_{1}}$ is given by

$$\begin{eqnarray}&&{{\varepsilon }_{1}}=\left( 2{{\partial }_{A}}{{\sigma }_{0}}{{\partial }^{A}}{{\sigma }_{1}}+\frac{2{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\omega }_{1}}}{\eta _{0}^{2}}-\frac{2{{\sigma }_{1}}|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}} \right)\;\rho .\end{eqnarray} \tag{ 143 }$$

Gauge equations: for the shift, we have

$$\begin{eqnarray}&&{{\left( \mathcal{L}{{\beta }_{1}} \right)}^{AB}}=2{{{\rm e}}^{-2{{u}_{0}}}}\rho \chi _{1}^{AB}.\end{eqnarray} \tag{ 144 }$$

The energy density ${{\varepsilon }_{2}}$ defined previously in equation (49) is given by

$$\begin{eqnarray}\begin{array}{rcl} {{\varepsilon }_{2}} & = & \left( 2\frac{{{{\rm e}}^{2{{u}_{0}}}}}{{{\rho }^{2}}}{{p}^{2}}+2\frac{{{{\rm e}}^{2{{u}_{0}}}}}{{{\rho }^{2}}\eta _{0}^{2}}{{d}^{2}}+4{{{\rm e}}^{-2{{u}_{0}}}}\chi _{1}^{AB}{{\chi }_{1AB}} \right. \\ {} & {} & \left. +2{{\left| \partial {{\sigma }_{1}} \right|}^{2}}+2\frac{|\partial {{\omega }_{1}}{{|}^{2}}}{\eta _{0}^{2}}+4\frac{|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}}\sigma _{1}^{2}-8\frac{{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\omega }_{1}}{{\sigma }_{1}}}{\eta _{0}^{2}} \right)\;\rho . \\ \end{array}\end{eqnarray} \tag{ 145 }$$

Note that the energy density in (145) is not positive definite, and hence it is by no means obvious that the energy m₂ is positive.

Theorem 4.1. Consider a smooth solution of the linearized equations presented earlier, such that it satisfies the fall-off decay conditions at infinity in (23), the decay conditions at the extra asymptotic end at the origin in (26), and the regularity conditions in (24) and (25) at the axis. Assume also that ω₁ satisfies the following conditions. At the axis, we have

$$\begin{eqnarray}&&{{\mathop{{\bar{\omega }}}\limits^{.}}_{1}},{{\bar{\omega }}_{1}}={{O}_{1}}(1),\end{eqnarray} \tag{ 146 }$$

and both at infinity and at the origin, we impose

$$\begin{eqnarray}&&{{\mathop{{\bar{\omega }}}\limits^{.}}_{1}},{{\bar{\omega }}_{1}}={{o}_{1}}({{r}^{-5/2}}),\end{eqnarray} \tag{ 147 }$$

where we have defined

$$\begin{eqnarray}&&{{\bar{\omega }}_{1}}=\frac{{{\omega }_{1}}}{\eta _{0}^{2}}.\end{eqnarray} \tag{ 148 }$$

Then, we have:

• (i)  
  The first order mass m₁ defined by (53) with ${{\varepsilon }_{1}}$ given by (143) vanishes ${{m}_{1}}=0$. The second order mass m₂ defined by (54) with ${{\varepsilon }_{2}}$ given by (145) is equal to the following expression, which is explicitly definite positive

  $$\begin{eqnarray}&&{{m}_{2}}=\frac{1}{16}{{\int }_{\mathbb{R}_{+}^{2}}}{{\bar{\epsilon }}_{2}}\;\rho {\rm d}\rho {\rm d}z,\end{eqnarray} \tag{ 149 }$$

  where

  $$\begin{eqnarray}&&\begin{array}{l} {{{\bar{\epsilon }}}_{2}}=\left( 2\frac{{{{\rm e}}^{2{{u}_{0}}}}}{{{\rho }^{2}}}{{p}^{2}}+2\frac{{{{\rm e}}^{2{{u}_{0}}}}}{{{\rho }^{2}}\eta _{0}^{2}}{{d}^{2}}+4{{e}^{-2{{u}_{0}}}}\chi _{1}^{AB}{{\chi }_{1AB}} \right. \\ \qquad +{{\left( \partial {{\sigma }_{1}}+{{\omega }_{1}}\eta _{0}^{-2}\partial {{\omega }_{0}} \right)}^{2}}+{{\left( \partial \left( {{\omega }_{1}}\eta _{0}^{-1} \right)-\eta _{0}^{-1}{{\sigma }_{1}}\partial {{\omega }_{0}} \right)}^{2}} \\ \quad \ \ \left. +{{\left( \eta _{0}^{-1}{{\sigma }_{1}}\partial {{\omega }_{0}}-{{\omega }_{1}}\eta _{0}^{-2}\partial {{\eta }_{0}} \right)}^{2}} \right)\;\rho . \\ \end{array}\end{eqnarray} \tag{ 150 }$$
• (ii)  
  The mass m₂ is conserved along the evolution.

Note that the boundary condition in (146) at the axis (outside the origin) is identical to the one used in Minkowski in section 3, since η₀ behaves like ρ² at the axis.

Proof. (i) We first prove that ${{m}_{1}}=0$. Take the density ${{\varepsilon }_{1}}$ given by (143); for the first term, we have

$$\begin{eqnarray}&&2\rho {{\partial }_{A}}{{\sigma }_{0}}{{\partial }^{A}}{{\sigma }_{1}}=2{{\partial }^{A}}\left( \rho {{\sigma }_{1}}{{\partial }_{A}}{{\sigma }_{0}} \right)-2{{\sigma }_{1}}{{\partial }^{A}}\left( \rho {{\partial }_{A}}{{\sigma }_{0}} \right)\end{eqnarray} \tag{ 151 }$$

$$\begin{eqnarray}&&=\;2{{\partial }^{A}}\left( \rho {{\sigma }_{1}}{{\partial }_{A}}{{\sigma }_{0}} \right)-2\rho \sigma _{1}^{(3)}\Delta {{\sigma }_{0}}\end{eqnarray} \tag{ 152 }$$

$$\begin{eqnarray}&&=\;2{{\partial }^{A}}\left( \rho {{\sigma }_{1}}{{\partial }_{A}}{{\sigma }_{0}} \right)-2\rho {{\sigma }_{1}}\frac{|\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}},\end{eqnarray} \tag{ 153 }$$

where in line (152) we have used the definition of $^{(3)}\Delta$ given by equation (6) and in line (153) we have used the stationary equation (133).

For the second term, we have

$$\begin{eqnarray}&&\frac{2\rho {{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\omega }_{1}}}{\eta _{0}^{2}}=2{{\partial }^{A}}\left( \frac{\rho {{\omega }_{1}}{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)-2{{\omega }_{1}}{{\partial }^{A}}\left( \frac{\rho {{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)\end{eqnarray} \tag{ 154 }$$

$$\begin{eqnarray}&&=2{{\partial }^{A}}\left( \frac{\rho {{\omega }_{1}}{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right),\end{eqnarray} \tag{ 155 }$$

where in line (155) we have used the stationary equation (134). Summing up these terms, we find

$$\begin{eqnarray}&&{{\varepsilon }_{1}}={{\partial }_{A}}{{t}^{A}},\end{eqnarray} \tag{ 156 }$$

where

$$\begin{eqnarray}&&{{t}_{A}}=2\rho {{\sigma }_{1}}{{\partial }_{A}}{{\sigma }_{0}}+2\rho {{\omega }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}}.\end{eqnarray} \tag{ 157 }$$

We integrate equation (156) in the domain shown in figure 2 for some finite δ and L with $0\lt \delta \lt L$. At the axis, the first term in (157) clearly vanishes. The second term also vanishes by the assumption in (146) and the behavior in (136) of the background quantities. Hence, the integral of (156) contains only the two boundary terms ${{C}_{\delta }}$ and C_L . Then, we take the limit $\delta \to 0$ and $L\to \infty$. Using the assumptions in (147) on ω₁, the assumptions (23) and (26) on σ₁, and the background fall-off in (135), we obtain that these two boundary integrals vanish. Hence, it follows that ${{m}_{1}}=0$.

We prove now the positivity of m₂. The proof is identical to the proof of positivity presented in section 3 of [18], which is based on the Carter identity [10]. The last four terms in (145) are identical to the integrand of equation (24) in [18] (in that reference, a different notation is used, namely ${{\sigma }_{1}}=\alpha$, ${{\omega }_{1}}=y$, ${{\eta }_{0}}=X$, and ${{\omega }_{0}}=Y$). Then, the Carter identity given by equation (57) in [18] in the notation of this article can be written as

$$\begin{eqnarray}&&{{\bar{\epsilon }}_{2}}-{{\epsilon }_{2}}={{\partial }_{A}}{{t}^{A}},\end{eqnarray} \tag{ 158 }$$

where

$$\begin{eqnarray}&&{{t}_{A}}=2\rho \left( 2{{\sigma }_{1}}{{\partial }_{A}}{{\sigma }_{1}}+{{\omega }_{1}}\frac{{{\partial }_{A}}{{\omega }_{1}}}{\eta _{0}^{2}}-2{{\sigma }_{1}}{{\omega }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}}+\frac{{{\omega }_{1}}}{{{\eta }_{0}}}{{\partial }_{A}}\left( \frac{{{\omega }_{1}}}{{{\eta }_{0}}} \right) \right),\end{eqnarray} \tag{ 159 }$$

and ${{\bar{\epsilon }}_{2}}$ is given by (150). Recall that the divergence term in the right side of equation (158) has two contributions: one is the right side of equation (57) in [18] and the other comes from the integration by parts in equation (63) in [18]. Also note that in [18] Cartesian coordinates in ${{\mathbb{R}}^{3}}$ are used for the integration, and here we use cylindrical coordinates, and hence the factor ρ appears in (159). Integrating equation (158) and using the fall-off conditions at infinity and at the axis, it follows that m₂ is given by (149), and hence it is positive.

(ii) To prove the conservation of m₂ we take a time derivative of the mass density (145), and we obtain

$$\begin{eqnarray}\begin{array}{rcl} {{{\dot{\varepsilon }}}_{2}} & = & 4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho }p\dot{p}+4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho \eta _{0}^{2}}d\dot{d}+8{{e}^{-2{{u}_{0}}}}\rho \chi _{1}^{AB}{{{\dot{\chi }}}_{1AB}} \\ {} & {} & +4\rho {{\partial }_{A}}{{\sigma }_{1}}{{\partial }^{A}}{{{\dot{\sigma }}}_{1}}+4\rho \frac{{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{{\dot{\omega }}}_{1}}}{\eta _{0}^{2}}-8\rho {{\sigma }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}}{{\partial }^{A}}{{{\dot{\omega }}}_{1}} \\ {} & {} & +8\rho \frac{|\partial {{\omega }_{0}}{{|}^{2}}{{\sigma }_{1}}{{{\dot{\sigma }}}_{1}}}{\eta _{0}^{2}}-8\rho {{{\dot{\sigma }}}_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}}{{\partial }^{A}}{{\omega }_{1}}. \\ \end{array}\end{eqnarray} \tag{ 160 }$$

The strategy is very similar (but the calculations are lengthier) than in the Minkowski case discussed in section 3: using the linearized equations, we will write the right side of (160) as a total divergence. We proceed analyzing term by term.

For the first two terms, we just use the definition of p and d given in equations (139) and (140), respectively. We obtain

$$\begin{eqnarray}&&4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho }p\dot{p}=4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho }\dot{p}\left( {{{\dot{\sigma }}}_{1}}-\frac{2\beta _{1}^{\rho }}{\rho }-\beta _{1}^{A}{{\partial }_{A}}{{\sigma }_{0}} \right),\end{eqnarray} \tag{ 161 }$$

$$\begin{eqnarray}&&4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho \eta _{0}^{2}}d\dot{d}=4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho \eta _{0}^{2}}\dot{d}\left( {{{\dot{\omega }}}_{1}}-\beta _{1}^{A}{{\partial }_{A}}{{\omega }_{0}} \right).\end{eqnarray} \tag{ 162 }$$

For the third term, we have

$$\begin{eqnarray}&&8{{{\rm e}}^{-2{{u}_{0}}}}\rho \chi _{1}^{AB}{{\dot{\chi }}_{1AB}}=8\dot{\chi }_{1}^{AB}{{\partial }_{A}}{{\beta }_{1B}}\end{eqnarray} \tag{ 163 }$$

$$\begin{eqnarray}&&=\;8{{\partial }_{A}}({{\beta }_{1B}}\dot{\chi }_{1}^{AB})-8{{\beta }_{1B}}{{\partial }_{A}}\dot{\chi }_{1}^{AB}\end{eqnarray} \tag{ 164 }$$

$$\begin{eqnarray}&&=\;8{{\partial }_{A}}({{\beta }_{1B}}\dot{\chi }_{1}^{AB})+4\frac{{{e}^{2{{u}_{0}}}}}{\rho }\left( \dot{p}\left( 2\frac{\beta _{1}^{\rho }}{\rho }+\beta _{1}^{A}{{\partial }_{A}}{{\sigma }_{0}} \right)+\dot{d}\frac{\beta _{1}^{A}{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right),\end{eqnarray} \tag{ 165 }$$

where in line (163) we have used equation (144) and the fact that χ₁^AB is trace free, and in line (165) we have used the time derivative of equation (141).

For the fourth term, we have

$$\begin{eqnarray}&&4\rho {{\partial }_{A}}{{\sigma }_{1}}{{\partial }^{A}}{{\dot{\sigma }}_{1}}=4{{\partial }_{A}}(\rho {{\dot{\sigma }}_{1}}{{\partial }^{A}}{{\sigma }_{1}})-4{{\dot{\sigma }}_{1}}{{\partial }^{A}}(\rho {{\partial }_{A}}{{\sigma }_{1}})\end{eqnarray} \tag{ 166 }$$

$$\begin{eqnarray}&&=\;4{{\partial }_{A}}(\rho {{\dot{\sigma }}_{1}}{{\partial }^{A}}{{\sigma }_{1}})-4{{\dot{\sigma }}_{1}}{{\rho }^{(3)}}\Delta {{\sigma }_{1}}\end{eqnarray} \tag{ 167 }$$

$$\begin{eqnarray}\begin{array}{cll} {} & = & 4{{\partial }_{A}}(\rho {{{\dot{\sigma }}}_{1}}{{\partial }^{A}}{{\sigma }_{1}})-4\frac{{{{\dot{\sigma }}}_{1}}\dot{p}{{{\rm e}}^{2{{u}_{0}}}}}{\rho } \\ {} & {} & +8\frac{{{{\dot{\sigma }}}_{1}}}{\eta _{0}^{2}}\rho {{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{\omega }_{0}}-8\frac{{{{\dot{\sigma }}}_{1}}{{\sigma }_{1}}\rho |\partial {{\omega }_{0}}{{|}^{2}}}{\eta _{0}^{2}}, \\ \end{array}\end{eqnarray} \tag{ 168 }$$

where in line (167) we used the definition of the operator $^{(3)}\Delta$ given by equation (6) and in line (168) we used equation (137).

For the fifth term, we obtain

$$\begin{eqnarray}&&4\rho \frac{{{\partial }_{A}}{{{\dot{\omega }}}_{1}}{{\partial }^{A}}{{\omega }_{1}}}{\eta _{0}^{2}}=4{{\partial }^{A}}\left( \frac{\rho {{{\dot{\omega }}}_{1}}{{\partial }_{A}}{{\omega }_{1}}}{\eta _{0}^{2}} \right)-4{{\dot{\omega }}_{1}}{{\partial }^{A}}\left( \frac{\rho {{\partial }_{A}}{{\omega }_{1}}}{\eta _{0}^{2}} \right)\end{eqnarray} \tag{ 169 }$$

$$\begin{eqnarray}&&=\;4{{\partial }^{A}}\left( \frac{\rho {{{\dot{\omega }}}_{1}}{{\partial }_{A}}{{\omega }_{1}}}{\eta _{0}^{2}} \right)-4\frac{\rho {{{\dot{\omega }}}_{1}}}{\eta _{0}^{2}}\left( ^{(3)}\Delta {{\omega }_{1}}-\frac{4{{\partial }_{\rho }}{{\omega }_{1}}}{\rho }-2{{\partial }_{A}}{{\omega }_{1}}{{\partial }^{A}}{{\sigma }_{0}} \right)\end{eqnarray} \tag{ 170 }$$

$$\begin{eqnarray}&&=\;4{{\partial }^{A}}\left( \frac{\rho {{{\dot{\omega }}}_{1}}{{\partial }_{A}}{{\omega }_{1}}}{\eta _{0}^{2}} \right)-4\frac{{{{\rm e}}^{2{{u}_{0}}}}}{\rho \eta _{0}^{2}}\dot{d}{{\dot{\omega }}_{1}}-8\frac{\rho {{{\dot{\omega }}}_{1}}{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\sigma }_{1}}}{\eta _{0}^{2}},\end{eqnarray} \tag{ 171 }$$

where in line (170) we have used the definition of the operator $^{(3)}\Delta$ given in equation (6) and the definition of η₀ given in equation (35). In line (171) we have used the evolution equation in (138).

For the sixth term, we obtain

$$\begin{eqnarray}&&-8\rho {{\sigma }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{{\dot{\omega }}}_{1}}}{\eta _{0}^{2}}=-8{{\partial }^{A}}\left( \rho {{{\dot{\omega }}}_{1}}{{\sigma }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)+8{{\dot{\omega }}_{1}}{{\partial }^{A}}\left( \rho {{\sigma }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)\end{eqnarray} \tag{ 172 }$$

$$\begin{eqnarray}&&=-8{{\partial }^{A}}\left( {{{\dot{\omega }}}_{1}}{{\sigma }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)+8\rho {{\dot{\omega }}_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\sigma }_{1}}}{\eta _{0}^{2}}+8{{\dot{\omega }}_{1}}{{\sigma }_{1}}{{\partial }^{A}}\left( \frac{\rho {{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)\end{eqnarray} \tag{ 173 }$$

$$\begin{eqnarray}&&=-8{{\partial }^{A}}\left( {{{\dot{\omega }}}_{1}}{{\sigma }_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}}{\eta _{0}^{2}} \right)+8\rho {{\dot{\omega }}_{1}}\frac{{{\partial }_{A}}{{\omega }_{0}}{{\partial }^{A}}{{\sigma }_{1}}}{\eta _{0}^{2}},\end{eqnarray} \tag{ 174 }$$

where in line (174) we have used the stationary equation in (134).

We sum the six terms obtained here plus the two last terms in (160), where only the divergence terms survive. We obtain

$$\begin{eqnarray}&&{{\dot{\varepsilon }}_{2}}={{\partial }_{A}}{{t}^{A}},\end{eqnarray} \tag{ 175 }$$

where

$$\begin{eqnarray}&&{{t}^{A}}=4\rho {{\dot{\sigma }}_{1}}{{\partial }^{A}}{{\sigma }_{1}}+8{{\beta }_{1B}}\dot{\chi }_{1}^{AB}+4\frac{\rho {{{\dot{\omega }}}_{1}}{{\partial }^{A}}{{\omega }_{1}}}{\eta _{0}^{2}}-8{{\dot{\omega }}_{1}}{{\sigma }_{1}}\frac{{{\partial }^{A}}{{\omega }_{0}}}{\eta _{0}^{2}}.\end{eqnarray} \tag{ 176 }$$

Remarkably, we get only one extra term compared with the Minkowski case (compare (176) with the sum of (108) and (118)).

We integrate equation (175) in the domain shown in figure 2. The boundary term at the axis vanished according to the hypothesis in (25). Then, we take the limit $\delta \to 0$ and $L\to \infty$, and the other two boundary integrals also vanished according to the fall-off conditions in (23), (26) and (146), (147). □

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This work was supported by grant PICT-2010–1387 of CONICET (Argentina) and grant Secyt-UNC (Argentina).

In this appendix we explicitly write the Kerr black hole metric in the maximal-isothermal gauge described in section 2. In particular, we show that in this gauge the metric satisfies the conditions in (29).

The Kerr metric, with parameters (m, a), in Boyer–Lindquist coordinates $(t,\tilde{r},\theta ,\phi )$ is given by

$$\begin{eqnarray}&&g=-V{\rm d}{{t}^{2}}+2W{\rm d}t{\rm d}\phi +\frac{\Sigma }{\Xi }{\rm d}{{\tilde{r}}^{2}}+\Sigma {\rm d}{{\theta }^{2}}+\eta {\rm d}{{\phi }^{2}},\end{eqnarray} \tag{ A.1 }$$

where

$$\begin{eqnarray}&&\Xi ={{\tilde{r}}^{2}}+{{a}^{2}}-2m\tilde{r},\qquad \quad \Sigma ={{\tilde{r}}^{2}}+{{a}^{2}}{{{\rm cos} }^{2}}\;\theta ,\end{eqnarray} \tag{ A.2 }$$

and

$$\begin{eqnarray}&&V=\frac{\Xi -{{a}^{2}}{{{\rm sin} }^{2}}\theta }{\Sigma },\end{eqnarray} \tag{ A.3 }$$

$$\begin{eqnarray}&&W=-\frac{2ma\tilde{r}\;{{{\rm sin} }^{2}}\theta }{\Sigma },\end{eqnarray} \tag{ A.4 }$$

$$\begin{eqnarray}&&\eta =\left( \frac{{{\left( {{\tilde{r}}^{2}}+{{a}^{2}} \right)}^{2}}-\Xi {{a}^{2}}{{{\rm sin} }^{2}}\theta }{\Sigma } \right){{{\rm sin} }^{2}}\theta .\end{eqnarray} \tag{ A.5 }$$

The angular momentum is given by

$$\begin{eqnarray}&&J=ma.\end{eqnarray} \tag{ A.6 }$$

The metric in (A.1) is stationary and axially symmetric because it has the following two Killing vectors

$$\begin{eqnarray}&&{{\xi }^{\mu }}={{\left( \frac{\partial }{\partial t} \right)}^{\mu }},\qquad \quad {{\eta }^{\nu }}={{\left( \frac{\partial }{\partial \phi } \right)}^{\nu }},\end{eqnarray} \tag{ A.7 }$$

where ${{\xi }^{\mu }}$ is timelike near infinity (i.e., outside the ergosphere) and ${{\eta }^{\mu }}$ is spacelike and it vanished at the axis. The scalars in (A.3), (A.4), and (A.5) are written in terms of the Killing vectors as follows:

$$\begin{eqnarray}&&V=-{{\xi }^{\mu }}{{\xi }^{\nu }}{{g}_{\mu \nu }},\qquad \quad \eta ={{\eta }^{\mu }}{{\eta }^{\nu }}{{g}_{\mu \nu }},\qquad \quad W={{\eta }^{\mu }}{{\xi }^{\nu }}{{g}_{\mu \nu }}.\end{eqnarray} \tag{ A.8 }$$

In particular, η is the square norm of the axial Killing vector ${{\eta }^{\mu }}$. In these equations, we are using four-dimensional indices $\mu ,\nu \cdots$.

The twist potential ω of the axial Killing vector ${{\eta }^{\mu }}$ is given by

$$\begin{eqnarray}&&\omega =2ma\left( {{{\rm cos} }^{3}}\theta -3{\rm cos} \theta \right)-\frac{2m{{a}^{3}}{\rm cos} \theta {{{\rm sin} }^{4}}\theta }{\Sigma }.\end{eqnarray} \tag{ A.9 }$$

The three-dimensional Lorenzian metric h on the quotient manifold (see equation (26) on [24], we follow the notation of that article) is defined by

$$\begin{eqnarray}&&\eta {{g}_{\mu \nu }}={{h}_{\mu \nu }}+{{\eta }_{\mu }}{{\eta }_{\nu }}.\end{eqnarray} \tag{ A.10 }$$

Using the explicit form of the Kerr metric in (A.1) and the Killing vector ${{\eta }^{\mu }}$ we obtain that h is given by

$$\begin{eqnarray}&&h=-(V\eta +{{W}^{2}}){\rm d}{{t}^{2}}+\frac{\eta \Sigma }{\Xi }{\rm d}{{\tilde{r}}^{2}}+\eta \Sigma {\rm d}{{\theta }^{2}}.\end{eqnarray} \tag{ A.11 }$$

For the Kerr metric, the following remarkable relation holds:

$$\begin{eqnarray}&&V\eta +{{W}^{2}}=\Xi {{{\rm sin} }^{2}}\;\theta .\end{eqnarray} \tag{ A.12 }$$

Using (A.12) we further simplify the expression for the metric h

$$\begin{eqnarray}&&h=-\Xi {{{\rm sin} }^{2}}\;\theta {\rm d}{{t}^{2}}+\frac{\eta \Sigma }{\Xi }{\rm d}{{\tilde{r}}^{2}}+\eta \Sigma {\rm d}{{\theta }^{2}}.\end{eqnarray} \tag{ A.13 }$$

This metric is static. The foliation t = constant has zero extrinsic curvature, and hence it is a maximal foliation. The shift of this foliation also vanished, so then the condition in (29) is satisfied. However, the coordinates $(\tilde{r},\theta )$ are not isothermal because they do not satisfy the condition in (2).

To introduce isothermal coordinates, we will assume that $m\geqslant |a|$ (i.e., the Kerr metric in (A.1) describes a black hole). Let r be defined as the positive root of the equation

$$\begin{eqnarray}&&\tilde{r}=r+m+\frac{{{m}^{2}}-{{a}^{2}}}{4r},\end{eqnarray} \tag{ A.14 }$$

that is

$$\begin{eqnarray}&&r=\frac{1}{2}\left( \tilde{r}-m+\sqrt{\Xi } \right).\end{eqnarray} \tag{ A.15 }$$

We have

$$\begin{eqnarray}&&d\tilde{r}=\frac{\sqrt{\Xi }}{r}dr.\end{eqnarray} \tag{ A.16 }$$

We define the cylindrical coordinates $(\rho ,z)$ in terms of the spherical coordinates $(r,\theta )$ by the standard formula

$$\begin{eqnarray}&&\rho =r{\rm sin} \theta ,\qquad \quad z=r{\rm cos} \theta .\end{eqnarray} \tag{ A.17 }$$

Then the metric h in the new coordinate system $(t,\rho ,z)$ is given by

$$\begin{eqnarray}&&h=-{{\alpha }^{2}}{\rm d}{{t}^{2}}+{{{\rm e}}^{2u}}({\rm d}{{\rho }^{2}}+{\rm d}{{z}^{2}}),\end{eqnarray} \tag{ A.18 }$$

where

$$\begin{eqnarray}&&\alpha =\sqrt{\Xi }{\rm sin} \theta =\rho \left( 1-\frac{({{m}^{2}}-{{a}^{2}})}{4{{r}^{2}}} \right),\end{eqnarray} \tag{ A.19 }$$

and

$$\begin{eqnarray}&&{{{\rm e}}^{2u}}=\frac{\eta \Sigma }{{{r}^{2}}}.\end{eqnarray} \tag{ A.20 }$$

The intrinsic metric of the t = constant of the slices is

$$\begin{eqnarray}&&q={{{\rm e}}^{2u}}\left( {\rm d}{{\rho }^{2}}+{\rm d}{{z}^{2}} \right).\end{eqnarray} \tag{ A.21 }$$

That is, the coordinates system is isothermal.

The function σ is defined in terms of the norm η by

$$\begin{eqnarray}&&{{{\rm e}}^{\sigma }}=\frac{\eta }{{{\rho }^{2}}},\end{eqnarray} \tag{ A.22 }$$

The function q is given by

$$\begin{eqnarray}&&{{{\rm e}}^{2q}}=\frac{{{{\rm sin} }^{2}}\theta \Sigma }{\eta },\end{eqnarray} \tag{ A.23 }$$

We have the relation

$$\begin{eqnarray}&&u=q+\sigma +{\rm log} \rho .\end{eqnarray} \tag{ A.24 }$$

Note that the lapse satisfies the maximal gauge condition

$$\begin{eqnarray}&&\Delta \alpha =0.\end{eqnarray} \tag{ A.25 }$$

In the extreme case $m=|a|$ and hence we have

$$\begin{eqnarray}&&\alpha =\rho .\end{eqnarray} \tag{ A.26 }$$

In this appendix we prove the following Sobolev-type estimate.

Lemma B.1. There exists a constant $C\gt 0$ such that for all $u\in C_{0}^{\infty }({{\mathbb{R}}^{n}})$, with $n\geqslant 3$, the following inequality holds

$$\begin{eqnarray}&&C{{\left( {{\int }_{{{\mathbb{R}}^{n}}}}\left( |{{\partial }^{k}}u{{|}^{2}}+|{{\partial }^{k-1}}u{{|}^{2}} \right)\;{\rm d}{{x}^{n}} \right)}^{1/2}}\geqslant \mathop{{\rm sup} }\limits_{x\in {{\mathbb{R}}^{n}}}|u(x)|,\end{eqnarray} \tag{ B.1 }$$

where $k\gt n/2$.

Proof. The estimate in (B.1) will be a consequence of the following two classical estimates. The first one is the Gagliardo–Nirenberg–Sobolev inequality: assume that $1\leqslant p\lt n$, then there exists a constant C, depending only on p and n, such that

$$\begin{eqnarray}&&{\| {u}\| _{{{L}^{q}}({{\mathbb{R}}^{n}})}}\leqslant C\| \partial {{u}\| _{{{L}^{p}}({{\mathbb{R}}^{n}})}},\end{eqnarray} \tag{ B.2 }$$

for all $u\in C_{0}^{\infty }({{\mathbb{R}}^{n}})$, where

$$\begin{eqnarray}&&q=\frac{pn}{n-p}.\end{eqnarray} \tag{ B.3 }$$

The second estimate is the Morreyʼs inequality: assume $n\lt p\leqslant \infty$, then there exists a constant depending only on p and n, such that

$$\begin{eqnarray}&&\mathop{{\rm sup} }\limits_{x\in {{\mathbb{R}}^{n}}}\left| u\left( x \right) \right|\leqslant C{{\| u\| }_{{{W}^{1,p}}({{\mathbb{R}}^{n}})}}.\end{eqnarray} \tag{ B.4 }$$

See [27] for an elementary and clear presentation of these inequalities and the functional spaces ${{L}^{p}}({{\mathbb{R}}^{n}})$, ${{W}^{1,p}}({{\mathbb{R}}^{n}})$ involved in them.

We first observe that the estimate in (B.2) can be iterated as follows:

$$\begin{eqnarray}&&{{\| u\| }_{{{L}^{{{p}_{k}}}}({{\mathbb{R}}^{n}})}}\leqslant C{{\| \partial }^{k}}{{u\| }_{{{L}^{p}}({{\mathbb{R}}^{n}})}},\end{eqnarray} \tag{ B.5 }$$

where $1\leqslant k\leqslant n/p$, $1\lt p$, and p_k is given by

$$\begin{eqnarray}&&{{p}_{k}}=\frac{pn}{n-pk}.\end{eqnarray} \tag{ B.6 }$$

To prove (B.5) we use induction in k. For k = 1 the inequality in (B.5) reduces to (B.2). Assume that (B.5) is valid for k. If ${{\partial }^{k+1}}u\in {{L}^{p}}({{\mathbb{R}}^{n}})$, then by (B.2) we obtain that ${{\partial }^{k}}u\in {{L}^{q}}({{\mathbb{R}}^{n}})$ with q given by

$$\begin{eqnarray}&&q=\frac{pn}{n-p}.\end{eqnarray} \tag{ B.7 }$$

By the inductive hypothesis we obtain that $u\in {{L}^{{{q}_{k}}}}({{\mathbb{R}}^{n}})$ with

$$\begin{eqnarray}&&{{q}_{k}}=\frac{qn}{n-qk}.\end{eqnarray} \tag{ B.8 }$$

We substitute (B.7) in (34) to obtain

$$\begin{eqnarray}&&{{q}_{k}}=\frac{pn}{n-(k+1)p}.\end{eqnarray} \tag{ B.9 }$$

And then the desired result is proved.

To prove (B.1), note that the left side of (B.1) implies that ${{\partial }^{k-1}}w,\;{{\partial }^{k-1}}u\in {{L}^{2}}({{\mathbb{R}}^{n}})$ where $w=\partial u$. Then, we apply the inequality in (B.5) for both w and u to obtain that $w,u\in {{L}^{p}}({{\mathbb{R}}^{n}})$, with p given by

$$\begin{eqnarray}&&p=\frac{2n}{n-2k+2}.\end{eqnarray} \tag{ B.10 }$$

By hypothesis $k\gt n/2$, then we obtain that $p\gt n$. Hence, we have proved that $u\in {{W}^{1,p}}({{\mathbb{R}}^{n}})$ with $p\gt n$. We use the Morrey inequality in (B.4), and the desired result follows.□