Conformal Scattering Theory for the Dirac Equation on Kerr Spacetime

Abstract

We establish the construction of a conformal scattering theory of the spin-1/2 massless Dirac equation on the Kerr spacetime by using the conformal geometric method and a pointwise logarithmic decay estimate for the massless Dirac field. In particular, our construction is valid on the exterior domains of Kerr black hole spacetimes in the full sub-extremal range $|a|<M$.

Appendices

Appendix A

1.1 Goursat Problem for the Spin Wave Equations on Kerr Spacetime

In this part, we extend the results of Hörmander [34] for the spin-1/2 wave equations. The results of Hörmander were extended for the scalar wave equation by Nicolas [20] with the following minor modifications: the ${\mathcal {C}}^1$-metric, the continuous coefficients of the derivatives of the first-order and the terms of order zero have locally $L^\infty$-coefficients. We refer [13, 19, 21,22,23,24, 26] for the applications of the generalized results of Hörmander to solve the Goursat problem for the massless spin, tensorial equations, linear and semilinear wave equations. Here we will show how the Goursat problem is valid for the spin wave equations in the future ${\mathcal {I}}^+({\mathcal {S}})$ of ${\mathcal {S}}$ in $\bar{{\mathcal {B}}}_I$ (recall that ${\mathcal {S}}$ is the spacelike hypersurface in $\bar{{\mathcal {B}}}_I$ such that it passes ${{\mathscr {I}}}^+$ strictly in the past of the support data).

Let P be a point in $\bar{{\mathcal {B}}}_I$; we cut off ${\mathcal {I}}^+({\mathcal {S}})$ by the future neighborhood ${\mathcal {V}}$ of P such that ${\mathcal {V}}$ does not intersect with the support of Goursat data and get ${\mathfrak {B}}={\mathcal {I}}^+({\mathcal {S}})/{\mathcal {V}}$. We extend $({\mathfrak {B}},{\widetilde{g}})$ onto a cylindrical globally hyperbolic spacetime $({\mathfrak {M}}={{\mathbb {R}}}_t\times S^3, {\mathfrak {g}})$, where ${\mathfrak {g}}|_{{\mathfrak {B}}} = \tilde{g}|_{\bar{{\mathcal {B}}}_I}$ and the part of null conformal boundary ${\mathfrak {H}}^+\cup {{\mathscr {I}}}^+$ inside ${\mathcal {I}}^+({\mathcal {S}})/{\mathcal {V}}$ is extended as a null hypersurface ${\mathcal {C}}$ that is the graph of a Lipschitz function over $S^3$ and the data by zero on the rest of the extended hypersurface.

We consider the Goursat problem of the following spin wave equation in the spacetime $({\mathfrak {M}} = {{\mathbb {R}}}_t\times S^3, {\mathfrak {g}})$:

$$\begin{aligned} {\left\{ \begin{array}{ll} {\widehat{\Box }}{\widetilde{\phi }}_Z &{}= 0,\\ {\widetilde{\phi }}_A|_{{\mathcal {C}}} &{}= {\widetilde{\psi }}_A|_{{\mathcal {C}}} \in {\mathcal {C}}^\infty _0({\mathcal {C}}, {{{\mathbb {S}}}}_A),\\ \nabla _{{\mathfrak {g}}}^{AA'}{\widetilde{\phi }}_A|_{{\mathcal {C}}} &{} = {\widetilde{\zeta }}^{A'}|_{{\mathcal {C}}} \in {\mathcal {C}}^\infty _0({\mathcal {C}}, {{{\mathbb {S}}}}^{A'}), \end{array}\right. } \end{aligned}$$

(47)

where the operator ${\widehat{\Box }}$ defined by (16) acts on the full spin fields. Note that we can replace ${\widehat{\Box }}$ by the two other spin wave operators ${\check{\Box }}$ and $\breve{\Box }$ defined in Equations (17) and (18), respectively.

Following [35] the spacetime $({\mathfrak {M}} = {{\mathbb {R}}}_t\times S^3, {\mathfrak {g}})$ is parallelizable, i.e. it admits a continuous global frame in the sense that the tangent space at each point has a basis. Therefore, we can choose a global spin-frame $\left\{ o,\iota \right\}$ for ${\mathfrak {M}}$ such that in this spin-frame the Newman–Penrose tetrad is $\mathcal {C}^{\infty }$. Projecting (47) on $\left\{ o,\iota \right\}$ (see the end of “Appendix 6.1” for the projection of the covariant derivative equation $\nabla _{{\mathfrak {g}}}^{AA'}\phi _A|_{{\mathcal {C}}}$) we get the scalar matrix form as follows

$$\begin{aligned} {\left\{ \begin{array}{ll} P{\widetilde{\Phi }} + L_1 {\widetilde{\Phi }} &{}= 0,\\ ({\widetilde{\Phi }}, \,\partial _t{\widetilde{\Phi }})|_{t=0} &{}= ({\widetilde{\Psi }}, \, \partial _t{\widetilde{\Psi }}) \in {\mathcal {C}}^\infty _0({\mathcal {C}})\times {\mathcal {C}}^\infty _0({\mathcal {C}}), \end{array}\right. } \end{aligned}$$

(48)

where

$$\begin{aligned} P= \left( \begin{matrix} \Box &{}&{}0\\ 0&{}&{}\Box \end{matrix} \right) \end{aligned}$$

is the $2\times 2$-matrix diagram,

$$\begin{aligned} {\widetilde{\Phi }} = \left( \begin{matrix} {\widetilde{\phi }}_0\\ {\widetilde{\phi }}_1 \end{matrix} \right) , \, {\widetilde{\Psi }} = \left( \begin{matrix} {\widetilde{\psi }}_0\\ {\widetilde{\psi }}_1 \end{matrix} \right) \end{aligned}$$

are the components of ${\widetilde{\phi }}_{A}$ and ${\widetilde{\psi }}^{A'}$, respectively, on the spin-frames $\left\{ o_A,\iota _A \right\}$ and $\left\{ o^{A'},\iota ^{A'} \right\}$, respectively, and

$$\begin{aligned} L_1 = \left( \begin{matrix} L_1^{00}&{}&{}L_1^{01}\\[3pt] L_1^{10}&{}&{} L_1^{11} \end{matrix} \right) \end{aligned}$$

is the $2\times 2$-matrix where the components are the operators that have the ${\mathcal {C}}^\infty$-coefficients

$$\begin{aligned} L_1^{ij} = b_0^{ij}\partial _t + b_\alpha ^{ij}\partial _\alpha + c^{ij}. \end{aligned}$$

We have that ${\mathfrak {g}}$ is a ${\mathcal {C}}^1$-metric, the terms consisting of the derivatives in first order in $L_1$ have continuous coefficients and the terms without derivatives have locally $L^\infty$-coefficients. Then, the Goursat problem for the $2\times 2$-matrix wave equation (48) is well posed in $({\mathfrak {M}}={{\mathbb {R}}}_t\times S^3,{\mathfrak {g}})$ by applying the results in [20, Theorem 3 and Theorem 4].

Theorem 5.1

For the initial data $({\widetilde{\psi }}_i,\partial _t{\widetilde{\psi }}_i) \in {{{\mathcal {C}}}}_0^{\infty }({\mathcal {C}})\times {{\mathcal {C}}}_0^\infty ({\mathcal {C}})$ for all $i=0,1$, the $2\times 2$-matrix equation (48); hence, the spin wave equation (47) has a unique solution ${\widetilde{\Phi }} = ({\widetilde{\phi }}_0,{\widetilde{\phi }}_1)$ satisfying

$$\begin{aligned} {\widetilde{\phi }}_i \in {{\mathcal {C}}}({{\mathbb {R}}};H^1(S^3))\cap {\mathcal {C}}^1({{\mathbb {R}}};L^2(S^3)) \; \text{ for } \text{ all } \; i = 0,1. \end{aligned}$$

Using the finite propagation speed the solution ${\widetilde{\Phi }}$ vanishes in ${\mathcal {I}}^+({\mathcal {S}})/{\mathfrak {B}}$ by local uniqueness and causality. Therefore, the Goursat problem has a unique smooth solution in the future of ${\mathcal {S}}$, that is, the restriction of ${\widetilde{\Phi }}$ to ${\mathfrak {B}}$.

1.2 A Pointwise Logarithmic Decay Estimate for Dirac Field on Slow Kerr Spacetime

There are some methods to prove the pointwise decay of the field equations such as scalar wave, Dirac, Maxwell, linearized gravity and spin Teukolsky equations on Kerr spacetime (see [31, 36,37,38,39,40,41,42]). The methods are based on the transformation equations to the wave (or spin wave) equations with potentials which decay sufficiently to establish the decay of solutions. In this section, we will apply and develop the previous results obtained by Moschidis [27] to establish a pointwise logarithmic decay estimate for the massless Dirac field’s spin components on slow Kerr spacetimes in the full sub-extremal range $|a|<M$.

First, by using the curvature spinors and spinor form of commutators as in Sects. 2.3 and 2.4 we can show that the origin Dirac field $\phi _A$ satisfies the following spin wave equation

$$\begin{aligned} 2\nabla _{ZA'}\nabla ^{AA'}\phi _A = {\check{\Box }} \phi _Z + X_{ZA}{^{NA}}\phi _N = 0, \end{aligned}$$

(49)

where ${\check{\Box }} \phi _Z = \varepsilon ^{AM}\nabla _{A'[Z}\nabla {_{M]}}{^{A'}}\phi _A$ and $X_{ZA}{^{NA}}\phi _N = \varepsilon ^{AM}\nabla _{A'(Z}\nabla {_{M)}}{^{A'}}\phi _A$.

We note that the components of the curvature spinor $X_{ZA}{^{NA}}$ can be calculated from the components of the Riemann curvature ${R_{ab}}^{cd}$ given in Appendix 6.3. We can verify that the components of ${R_{ab}}^{cd}$ are equivalent to $r^{-3}$. Therefore, the coefficients of $X_{ZA}{^{NA}}$ are also equivalent to $r^{-3}$. On the other hand, by using the formulas (54) and (55) we can transform the spin wave operator ${\check{\Box }} \phi _Z$ (where ${\check{\Box }}$ acts on the full spinor field) to the scalar wave operator $\Box _g\phi _Z$ (where $\Box _g$ acts on the functional coefficients). This yields

$$\begin{aligned} {\check{\Box }} \phi _Z = \Box _g\phi _Z + \left\langle r\right\rangle ^{-2}\partial _\alpha \phi _Z + \left\langle r\right\rangle ^{-3}\phi _Z, \end{aligned}$$

where we denote the functions equivalent to r by $\left\langle r\right\rangle$. The term $\left\langle r\right\rangle ^{-2}\partial _\alpha \phi _Z + \left\langle r\right\rangle ^{-3}\phi _Z$ arises from the asymptotic part of Kerr metric (1) (which contains the angular momentum a and the mass M) to Minkowski metric.

Therefore, the spin wave equation (49) can be re-written by the following equivalence form

$$\begin{aligned} \Box _g\phi _Z = \left\langle r\right\rangle ^{-2}\partial _\alpha \phi _Z + \left\langle r\right\rangle ^{-3}\phi _Z. \end{aligned}$$

Projecting the above equation on the global spin-frame $\left\{ o_A,\iota _A \right\}$ on Block ${\mathcal {B}}_I$, we get the scalar matrix form as follows

$$\begin{aligned} P\Phi = \left\langle r\right\rangle ^{-2}\partial _\alpha \Phi + \left\langle r\right\rangle ^{-3}\Phi , \end{aligned}$$

(50)

where

$$\begin{aligned} P= \left( \begin{matrix} \Box &{}&{}0\\ 0&{}&{}\Box \end{matrix} \right) \end{aligned}$$

is the $2\times 2$-matrix scalar wave operator and $\Phi = \left( \begin{matrix} \phi _0\\ \phi _1 \end{matrix} \right)$ is the components of $\phi _{A}$ on the spin-frames $\left\{ o_A,\iota _A \right\}$. Therefore, Eq. (50) is equivalent to

$$\begin{aligned} \Box _g\phi _1 = \left\langle r\right\rangle ^{-2}\partial _\alpha \phi _1 + \left\langle r\right\rangle ^{-3}\phi _1 \end{aligned}$$

(51)

and

$$\begin{aligned} \Box _g\phi _0 = \left\langle r\right\rangle ^{-2}\partial _\alpha \phi _0 + \left\langle r\right\rangle ^{-3}\phi _0. \end{aligned}$$

(52)

Remark 6

Note that the Kerr metric in the full sub-extremal range $|a|<M$ satisfies the generalized conditions of asymptotic flat spacetimes in Sections 3.1, 7.1 and 8.1 in [43] and in Section 1.3 in [27]. Moreover, the coefficients in $L^1$ operator (which consists of the derivative terms of order less than or equal to one) in the right-hand sides of (51) and (52) decay sufficiently as $r\rightarrow \infty$ (for example, we can see that the expressions of Eqs. (51) and (52) in the coordinates $(t{^*},\, {{^*}t},\, \omega )$ have the same forms of the scalar wave equations (3.14) and (3.15) in [43]). Therefore, we can apply the previous results of the scalar wave equation $\Box _g\phi = 0$ obtained in [27, 43] and also [37] to Eqs. (51) and (52) on Kerr spacetimes in the full sub-extremal range $|a|<M$.

First, we apply the previous results (see [37, Theorems 3.1, 3.2 and Proposition 4.5.2]) to obtain the boundedness and integrated local energy estimates for the solutions of Eqs. (51) and (52) at the original and higher order in sub-extremal Kerr spacetime. Moreover, the $r^p$-weighted energy method can be also applied to Eqs. (51) and (52) as in Theorems 1.1 and 1.2 in [43]. Therefore, we can derive the polynomial decay estimates for the solutions of Eqs. (51) and (52) in the sub-extremal Kerr spacetime as in Corollary 3.1 in [37] or Theorems 1.3 and 1.4 in [43]. However, since we have discussed in Sect. 3.2 (then proved in Theorem 2) that we need not an improved decay to construct a conformal scattering theory for the massless Dirac field, we will provide here a weak decay that is sufficient for our construction.

In particular, we use the results in [27] to obtain a pointwise logarithmic decay estimate for the solutions of Eqs. (51) and (52). The foliation $\left\{ {\mathcal {H}}_T \right\} _T$ (given by (24)) satisfies the class of hyperboloidal hypersurfaces $\left\{ {\mathcal {S}}_t\right\} _{t}$ terminating at ${{\mathscr {I}}}^+$ in [27] (in detail, see [27, Equation (2.7)]) with the foliated time function $t = T + h(r) - r_*$. Applying Corollary 2.2 in [27], we have the logarithmic decay of the energy through ${\mathcal {H}}_T$ as follows: for any $m\in {{\mathbb {N}}}$, there exists a positive constant $C_m>0$ such that every smooth solutions of Eqs. (51) and (52) satisfy the following estimates

$$\begin{aligned} \int _{{\mathcal {H}}_T} J^N_{\mu }(\phi _i)n^\mu _{{\mathcal {H}}_T} \le \frac{C_m}{[\log (2+T+h(r)-r_*)]^{2m}}\left( \int _{\Sigma _0} r^{\delta _0}J^N_\mu (\phi _i)n^\mu + \sum _{j=0}^m\int _{\Sigma _0}J^N_\mu (T^j\phi _i)n^\mu \right) , \end{aligned}$$

where $i\in \left\{ 0,\, 1\right\} ,\, 0<\delta _0 \le 1$. Here, $T=\partial _t$ is the smoothing Killing vector field, N is red-shift vector field (see [27, Subsection 2.1.2]) and $J^N_\mu$ denotes the current of Eqs. (51) and (52) (see [27, Section 3.8]), $n^\mu _{{\mathcal {H}}_T}$ (resp. $n^\mu$) being the future directed normal to ${\mathcal {H}}_T$ (resp. $\Sigma _0$).

The above inequalities hold also for the higher order local energies of $\phi _i$ since the boundedness and integrated local energy estimates are valid for the high order energies of $\phi _i$. Using this fact, standard elliptic estimates and the Sobolev embedding theorem, we can derive a pointwise logarithmic decay estimate for $\phi _i$ as follows (see also Equation (2.14) in [27]):

$$\begin{aligned} \sup _{{\mathcal {H}}_T}|\phi _i|^2\le & {} \frac{C_m}{[\log (2+T+h(r)-r_*)]^{2m}} \sum _{k=0}^1\left( \int _{\Sigma _0} r^{\delta _0}J^N_\mu (N^k\phi _i)n^\mu \right. \\&\left. \quad +\, \sum _{j=0}^m\int _{\Sigma _0}J^N_\mu (N^kT^j\phi _i)n^\mu \right) , \end{aligned}$$

where $i=0,\, 1$.

Appendix B

1.1 Detailed Calculations for the Goursat Problem

The expression of the spinor field $\Xi ^{A'}$ on the spin-frame $\left\{ {\widetilde{o}},\,{\widetilde{\iota }} \right\}$ is as follows

$$\begin{aligned} \Xi ^{A'}= & {} \Xi ^{1'} {\widetilde{o}}^{A'} - \Xi ^{0'} {\widetilde{\iota }}^{A'}. \end{aligned}$$

The covariant derivative ${\widetilde{\nabla }}_{ZA'}$ acts on the full spinor field can be decomposed as

$$\begin{aligned} {\widetilde{\nabla }}_a \Xi = ({\widetilde{D}}\Xi ){\widetilde{n}}_a + ({\widetilde{D}}'\Xi ){\widetilde{l}}_a - ({\widetilde{\delta }}\Xi )\bar{{{{\widetilde{m}}}}}_a - ({\widetilde{\delta }}'\Xi ){\widetilde{m}}_a. \end{aligned}$$

We recall that the twelve values of the rescaled spin coefficients are

$$\begin{aligned} {\widetilde{\kappa }}= & {} {\widetilde{\sigma }}={\widetilde{\lambda }}={\widetilde{\nu }}=0,\\ {\widetilde{\tau }}= & {} -\frac{ia\sin \theta r}{\sqrt{2} \rho ^2} , \, {\widetilde{\pi }}=\frac{ia\sin \theta r}{\sqrt{2} {{\bar{p}}}^2}, \, {\widetilde{\rho }} = -\frac{iar\cos \theta }{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}, \, {\widetilde{\mu }} = \left( R-\frac{1}{{{\bar{p}}}}\right) \sqrt{\frac{\Delta }{2\rho ^2}},\\ {\widetilde{\varepsilon }}= & {} \frac{Mr^4 - a^2r^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}},\, {\widetilde{\alpha }} = \frac{r}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) , \\ {\widetilde{\beta }}= & {} \frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) , \\ {\widetilde{\gamma }}= & {} \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \left( \frac{ia\cos \theta }{\rho ^2}+R\right) \sqrt{\frac{\Delta }{2\rho ^2}}. \end{aligned}$$

The covariant derivative acts on the spin-frame $\left\{ {\widetilde{o}}_A,\, {\widetilde{\iota }}_A \right\}$ as (see Equation (4.5.26) in [28, Vol. 1]):

$$\begin{aligned}&{\widetilde{D}}{\widetilde{o}}_A = {\widetilde{\varepsilon }} {\widetilde{o}}_A - {\widetilde{\kappa }}{\widetilde{\iota }}_A = \frac{Mr^4 - a^2r^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}{\widetilde{o}}_A ,\\&{\widetilde{D}}{\widetilde{\iota }}_A = -{\widetilde{\varepsilon }}{\widetilde{\iota }}_A + {\widetilde{\pi }}{\widetilde{o}}_A = - \frac{Mr^4 - a^2r^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}{\widetilde{\iota }}_A + \frac{ia\sin \theta r}{\sqrt{2} {{\bar{p}}}^2} {\widetilde{o}}_A,\\&{\widetilde{\delta }}'{\widetilde{o}}_A = {\widetilde{\alpha }} {\widetilde{o}}_A - {\widetilde{\rho }}{\widetilde{\iota }}_A = \frac{r}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{o}}_A + \frac{iar\cos \theta }{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}{\widetilde{\iota }}_A,\\&{\widetilde{\delta }}'{\widetilde{\iota }}_A = -{\widetilde{\alpha }}{\widetilde{\iota }}_A + {\widetilde{\lambda }}{\widetilde{o}}_A = -\frac{r}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{\iota }}_A,\\&{\widetilde{\delta }}{\widetilde{o}}_A = {\widetilde{\beta }} {\widetilde{o}}_A - {\widetilde{\sigma }}{\widetilde{\iota }}_A = \frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{o}}_A,\\&{\widetilde{\delta }}{\widetilde{\iota }}_A = -{\widetilde{\beta }}{\widetilde{\iota }}_A + {\widetilde{\mu }}\tilde{o}_A = -\frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{\iota }}_A + \left( R-\frac{1}{{{\bar{p}}}}\right) \sqrt{\frac{\Delta }{2\rho ^2}}{\widetilde{\iota }}_A,\\&{\widetilde{D}}'{\widetilde{o}}_A {=} {\widetilde{\gamma }} {\widetilde{o}}_A {-} {\widetilde{\tau }}{\widetilde{\iota }}_A {=} \left( \frac{Mr^2 {-} a^2(r\sin ^2\theta {+} M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} {-} \left( \frac{ia\cos \theta }{\rho ^2}+R\right) \sqrt{\frac{\Delta }{2\rho ^2}}\right) {\widetilde{o}}_A -\frac{ia\sin \theta r}{\sqrt{2} \rho ^2}{\widetilde{\iota }}_A, \\&{\widetilde{D}}'{\widetilde{\iota }}_A = -{\widetilde{\gamma }}{\widetilde{\iota }}_A + {\widetilde{\nu }}{\widetilde{o}}_A = -\left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \left( \frac{ia\cos \theta }{\rho ^2}+R\right) \sqrt{\frac{\Delta }{2\rho ^2}}\right) {\widetilde{\iota }}_A. \end{aligned}$$

Similarly on the dual conjugation spin-frame $\left\{ {\widetilde{o}}^{A'}, \, {\widetilde{\iota }}^{A'} \right\}$ we have

$$\begin{aligned}&{\widetilde{D}}{\widetilde{o}}^{A'} = \frac{Mr^4 - a^2r^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}{\widetilde{o}}^{A'} ,\\&{\widetilde{D}}{\widetilde{\iota }}^{A'} = - \frac{Mr^4 - a^2r^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}{\widetilde{\iota }}^{A'} + \frac{ia\sin \theta r}{\sqrt{2} {{\bar{p}}}^2} {\widetilde{o}}^{A'},\\&{\widetilde{\delta }}'{\widetilde{o}}^{A'} = \frac{r}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{o}}^{A'} + \frac{iar\cos \theta }{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}{\widetilde{\iota }}^{A'},\\&{\widetilde{\delta }}'{\widetilde{\iota }}^{A'} = -\frac{r}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{\iota }}^{A'},\\&{\widetilde{\delta }}{\widetilde{o}}^{A'} = \frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{o}}^{A'},\\&{\widetilde{\delta }}{\widetilde{\iota }}^{A'} = -\frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) {\widetilde{\iota }}^{A'} + \left( R-\frac{1}{{{\bar{p}}}}\right) \sqrt{\frac{\Delta }{2\rho ^2}}{\widetilde{\iota }}^{A'},\\&{\widetilde{D}}'{\widetilde{o}}^{A'} = \left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \left( \frac{ia\cos \theta }{\rho ^2}+R\right) \sqrt{\frac{\Delta }{2\rho ^2}}\right) {\widetilde{o}}^{A'} -\frac{ia\sin \theta r}{\sqrt{2} \rho ^2}{\widetilde{\iota }}^{A'}, \\&{\widetilde{D}}'{\widetilde{\iota }}^{A'} = -\left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \left( \frac{ia\cos \theta }{\rho ^2}+R\right) \sqrt{\frac{\Delta }{2\rho ^2}}\right) {\widetilde{\iota }}^{A'}. \end{aligned}$$

We have the detailed expression of ${\widetilde{\nabla }}_{ZA'}\Xi ^{A'}$ given by

$$\begin{aligned}&{\widetilde{\nabla }}_{ZA'}\Xi ^{A'} = ({\widetilde{D}}\Xi ^{A'}){\widetilde{n}}_a + ({\widetilde{D}}'\Xi ^{A'}){\widetilde{l}}_a - ({\widetilde{\delta }}\Xi ^{A'})\bar{{{{\widetilde{m}}}}}_a - ({\widetilde{\delta }}'\Xi ^{A'}){\widetilde{m}}_a \nonumber \\&\quad = {\widetilde{D}} \left( \Xi ^{1'}{\widetilde{o}}^{A'} \right) {\widetilde{\iota }}_A{\widetilde{\iota }}_{A'} - {\widetilde{D}}' \left( \Xi ^{0'}{\widetilde{\iota }}^{A'} \right) {\widetilde{o}}_A {\widetilde{o}}_{A'} + {\widetilde{\delta }} \left( \Xi ^{0'} {\widetilde{\iota }}^{A'} \right) {\widetilde{\iota }}_{A} {\widetilde{o}}_{A'} - {\widetilde{\delta }}' \left( \Xi ^{1'}{\widetilde{o}}^{A'} \right) {\widetilde{o}}_A {\widetilde{\iota }}_{A'} \nonumber \\&\quad = \left\{ \!\left( \!-{\widetilde{D}} {-} \frac{Mr^4 {-} a^2r^2(r\sin ^2\theta {+} M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} \!\right) \Xi ^{1'} {+}\!\! \left( \!{\widetilde{\delta }} {-} \frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}{+}\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\!\right) \! \right) \Xi ^{0'}\!\right\} {\widetilde{\iota }}_A \nonumber \\&\qquad +\, \left\{ \left( -{\widetilde{D}}' + \left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \left( \frac{ia\cos \theta }{\rho ^2}+R\right) \sqrt{\frac{\Delta }{2\rho ^2}}\right) \right) \Xi ^{0'}\right. \nonumber \\&\left. \qquad +\, \left( {\widetilde{\delta }}' + \frac{r}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) \right) \Xi ^{1'} \right\} {\widetilde{o}}_A. \end{aligned}$$

(53)

Taking the restriction of the system (53) on ${{\mathscr {I}}}^+$ with noting that $\Xi ^{1'}|_{{{\mathscr {I}}}^+} =0$, we get only the restriction of the second equation

$$\begin{aligned} -\lim _{r\rightarrow \infty }{\widetilde{D}}'\Xi ^{0'}|_{{{\mathscr {I}}}^+} = -\lim _{r\rightarrow \infty }\sqrt{\frac{2(a^2+r^2)}{\Delta }} \partial _{{^*}t} \Xi ^{0'}|_{{{\mathscr {I}}}^+} = \sqrt{2}\partial _{{^*}t} \Xi ^{0'}|_{{{\mathscr {I}}}^+}= 0. \end{aligned}$$

Integrating these equations along ${{\mathscr {I}}}^+$, we get $\Xi ^{0'}|_{{{\mathscr {I}}}^+} = \; constant$. This leads to a fact that the Cauchy problem with the initial condition $\Xi ^{0'}|_{{{{\mathcal {V}}}}(P)\cap {{\mathscr {I}}}^+} = 0$ has a unique solution and it equals to zero.

By the same way, we can obtain the restriction of the rescaled equation on ${\mathfrak {H}}^+$ by considering the rescaled equation ${\widehat{\nabla }}_{ZA'}\Xi ^{A'}=0$ on the rescaled Kerr–star coordinates $(t{^*},R,\theta ,\varphi {^*})$. By the same calculations, we get the restriction on ${\mathfrak {H}}^+$ of ${\widehat{\nabla }}_{ZA'}\Xi ^{A'}=0$ which is the restriction of the first equation of (53) on ${\mathfrak {H}}^+$ with $\widetilde{.}$ replaced by $\widehat{.}$:

$$\begin{aligned}&\left( -{\widehat{D}} - \frac{Mr^4 - a^2r^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} \right) \Xi ^{1'} \\&\quad +\, \left( {\widehat{\delta }} - \frac{r}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) \right) \Xi ^{0'} = 0. \end{aligned}$$

Multiplying the above equation by $\dfrac{\sqrt{\Delta \rho ^2}}{\sqrt{2}(r_+^2+a^2)}$ and taking the restriction of the equation obtained on ${\mathfrak {H}}^+$ with noting that $\Xi ^{0'}|_{{\mathfrak {H}}^+} =0$, we get

$$\begin{aligned} \left( -\partial _{t{^*}} - \frac{a}{r_+^2 + a^2}\partial _{\varphi {^*}} - \frac{Mr_+^4 - a^2r_+^2(r_+\sin ^2\theta + M\cos ^2\theta )}{4(r_+^2+a^2)(r_+^2+ a^2\cos ^2\theta )} \right) \Xi ^{1'}|_{{\mathfrak {H}}^+} = 0. \end{aligned}$$

Using $\Delta |_{{\mathfrak {H}}^+} = r_+^2 - 2Mr_+ + a^2 = 0$ we obtain that

$$\begin{aligned} \left( -\partial _{t{^*}} - \frac{a}{2Mr_+}\partial _{\varphi {^*}} - \frac{r_+^2-Mr_+}{8M} \right) \Xi ^{1'}|_{{\mathfrak {H}}^+} = 0. \end{aligned}$$

Putting $v= t{^*}+ \frac{a}{2Mr_+}\varphi {^*}$, we have $\partial _v= \partial _{t{^*}} + \frac{a}{2Mr_+}\partial _{\varphi {^*}}$ and

$$\begin{aligned} \partial _v\Xi ^{1'}|_{{\mathfrak {H}}^+} + \frac{r_+^2-Mr_+}{8M} \Xi ^{1'}|_{{\mathfrak {H}}^+} = 0. \end{aligned}$$

By solving this equation, we get

$$\begin{aligned} \Xi ^{1'}|_{{\mathfrak {H}}^+} = Ce^{-\frac{r_+^2 - Mr_+}{8M}v}. \end{aligned}$$

From the initial condition $\Xi ^{1'}|_{{\mathcal {V}}(P)\cap {\mathfrak {H}}^+} = 0$, we find that $C=0$ and $\Xi ^{1'}|_{{\mathfrak {H}}^+}=0$. Therefore, we conclude that if the support of $\Xi ^{1'}|_{{\mathfrak {H}}^+}$ is compact and far away from $i^+$, then the rescaled equation ${\widehat{\nabla }}_{ZA'}\Xi ^{A'}=0$ leads to $\Xi ^{1'}|_{{\mathfrak {H}}^+} = 0$.

1.2 Spin Coefficients and Derivations of the Origin Spin-Frame

Using the Newman–Penrose tetrad normalization (6), the spin coefficients are calculated (see [12]):

$$\begin{aligned} \kappa= & {} {\tilde{\sigma }} = \lambda = \nu = 0,\\ \tau= & {} -\frac{ia\sin \theta }{\sqrt{2} \rho ^2} , \, \pi = \frac{ia\sin \theta }{\sqrt{2} {{\bar{p}}}^2}, \, {\tilde{\rho }} = \mu = -\frac{1}{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}, \, \varepsilon = \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}},\\ \alpha= & {} \frac{1}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{\bar{p}}}-\frac{\cot \theta }{2} + \frac{a^2\sin \theta \cos \theta }{2\rho ^2} \right) , \, \beta = \frac{1}{\sqrt{2}p} \left( \frac{\cot \theta }{2} +\frac{a^2\sin \theta \cos \theta }{2\rho ^2} \right) ,\\ \gamma= & {} \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \frac{ia\cos \theta }{\rho ^2}\sqrt{\frac{\Delta }{2\rho ^2}}; \end{aligned}$$

here we use ${\tilde{\sigma }}$ and ${\tilde{\rho }}$ to avoid the confusion with the parameters $\sigma$ and $\rho$ in the Kerr metric.

The covariant derivative acts on the spin-frame $\left\{ o_A,\, \iota _A \right\}$ as (see Equation (4.5.26) in [28, Vol. 1]):

$$\begin{aligned} \begin{aligned} Do_A&= \varepsilon o_A - \kappa \iota _A = \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}o_A,\\ D\iota _A&= -\varepsilon \iota _A + \pi o_A = -\frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}\iota _A + \frac{ia\sin \theta }{\sqrt{2} {{\bar{p}}}^2} o_A,\\ \delta 'o_A&= \alpha o_A - {\tilde{\rho }}\iota _A = \frac{1}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) o_A + \frac{1}{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}\iota _A,\\ \delta '\iota _A&= -\alpha \iota _A + \lambda o_A = -\frac{1}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) \iota _A,\\ \delta o_A&= \beta o_A - {\tilde{\sigma }}\iota _A = \frac{1}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) o_A,\\ \delta \iota _A&= -\beta \iota _A + \mu o_A = -\frac{1}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) \iota _A - \frac{1}{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}\iota _A,\\ D'o_A&= \gamma o_A - \tau \iota _A = \left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \frac{ia\cos \theta }{\rho ^2}\sqrt{\frac{\Delta }{2\rho ^2}}\right) o_A + \frac{ia\sin \theta }{\sqrt{2} \rho ^2}\iota _A, \\ D'\iota _A&= -\gamma \iota _A + \nu o_A = -\left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \frac{ia\cos \theta }{\rho ^2}\sqrt{\frac{\Delta }{2\rho ^2}}\right) \iota _A. \end{aligned} \end{aligned}$$

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Similarly on the dual conjugation spin-frame $\left\{ o^{A'}, \, \iota ^{A'} \right\}$ we have

$$\begin{aligned} \begin{aligned} Do^{A'}&= \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} o^{A'} ,\\ D\iota ^{A'}&= -\frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}}\iota ^{A'} + \frac{ia\sin \theta }{\sqrt{2} {{\bar{p}}}^2} o^{A'},\\ \delta 'o^{A'}&= \frac{1}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) o^{A'} + \frac{1}{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}\iota ^{A'},\\ \delta '\iota ^{A'}&= -\frac{1}{\sqrt{2}{\bar{p}}}\left( \frac{ia\sin \theta }{{{\bar{p}}}}-\frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) \iota ^{A'},\\ \delta o^{A'}&= \frac{1}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) o^{A'},\\ \delta \iota ^{A'}&= -\frac{1}{\sqrt{2} p}\left( \frac{\cot \theta }{2}+\frac{a^2\sin \theta \cos \theta }{2\rho ^2}\right) \iota ^{A'} + - \frac{1}{{{\bar{p}}}}\sqrt{\frac{\Delta }{2\rho ^2}}\iota ^{A'},\\ D'o^{A'}&= \left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \frac{ia\cos \theta }{\rho ^2}\sqrt{\frac{\Delta }{2\rho ^2}}\right) o^{A'} + \frac{ia\sin \theta }{\sqrt{2} \rho ^2}\iota ^{A'}, \\ D'\iota ^{A'}&= -\left( \frac{Mr^2 - a^2(r\sin ^2\theta + M\cos ^2\theta )}{2\rho ^2\sqrt{2\Delta \rho ^2}} - \frac{ia\cos \theta }{\rho ^2}\sqrt{\frac{\Delta }{2\rho ^2}}\right) \iota ^{A'}. \end{aligned} \end{aligned}$$

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1.3 Riemann Curvature Tensor of Kerr Metric

With the usual coordinate transformation $c=\cos \theta$, the Kerr metric (1) becomes

$$\begin{aligned} g= & {} \left( 1-\frac{2Mr}{\rho ^2}\right) \mathrm {d}t^2 +\frac{4aMr(1-c^2)}{\rho ^2}\mathrm {d}t\mathrm {d}\varphi -\frac{\rho ^2}{\Delta }\mathrm {d}r^2- \frac{\rho ^2}{1-c^2}\mathrm {d}c^2\\&\quad -\frac{\sigma ^2}{\rho ^2}(1-c^2)\mathrm {d}\varphi ^2, \end{aligned}$$

where $\rho ^2=r^2+a^2c^2$, $\Delta =r^2-2Mr+a^2$ and $\sigma ^2 = (r^2+a^2)\rho ^2+ 2Mra^2(1-c^2)$.

The nonzero components of the Riemann curvature tensor are

$$\begin{aligned} R_{r c r c}= & {} \frac{3 a^2c^2Mr - Mr^3}{(1 - c^2)\rho ^2\Delta },\, R_{rc \varphi t} = - \frac{ac(a^2c^2M - 3Mr^2)}{\rho ^4},\\ R_{r \varphi r \varphi }= & {} -\frac{(1 - c^2)}{\rho ^6 \Delta }(- 9 a^6c^2Mr + 6a^6c^4Mr + 12a^4c^2M^2r^2 - 12a^4c^4M^2r^2)\\&-\, \frac{(1 - c^2)}{\rho ^6 \Delta }(3 a^4Mr^3 - 14a^4c^2Mr^3 + 6a^4c^4Mr^3 - 4a^2M^2r^4)\\&-\, \frac{(1 - c^2)}{\rho ^6 \Delta }( + 4a^2c^2M^2r^4 + 4a^2Mr^5 - 5a^2c^2Mr^5 + Mr^7),\\ R_{r\varphi r t}= & {} \frac{a (1 - c2) (9a^4c^2Mr - 12 a^2c^2M^2r^2 - 3a^2Mr^3 + 9a^2c^2Mr^3 + 4M^2r^4 - 3Mr^5)}{\rho ^6 \Delta },\\ R_{r\varphi c \varphi }= & {} \frac{3a^2c(1 - c^2) (a^2 + r^2) (a^2c^2M - 3Mr^2)}{\rho ^6},\\ R_{r\varphi c t}= & {} -\frac{ac(- 3a^2 + 2a^2c^2 - r^2) (a^2c^2M - 3Mr^2)}{\rho ^6},\\ R_{rtrt}= & {} -\frac{(-9a^4c^2Mr {+} 3 a^4c^4Mr {+} 12a^2c^2M^2r^2 {+} 3a^2Mr^3 {-} 7a^2c^2Mr^3 {-} 4M^2r^4 {+} 2Mr^5)}{\rho ^6\Delta },\\ R_{rtc\varphi }= & {} -\frac{ac(-3a^2 + a^2c^2 - 2r^2) (a^2c^2M - 3Mr^2)}{\rho ^6},\, R_{rtct}= \frac{3a^2c(a^2c^2M -3Mr^2)}{\rho ^6},\\ R_{c \varphi c \varphi }= & {} \frac{9a^6c^2Mr + 3a^6c^4Mr + 6a^4c^2M^2r^2 - 6 a^4c^4M^2r^2 + 3a^4Mr^3 - 16a^4c^2Mr^3}{\rho ^6} \\&+\, \frac{3a^4c^4Mr^3 - 2a^2M^2r^4+ 2a^2c^2M^2r^4 + 5a^2Mr^5 - 7a^2c^2Mr^5 + 2Mr^7}{\rho ^6},\\ R_{c \varphi ct}= & {} -\frac{a(9a^4c^2Mr - 6a^2c^2M^2r^2 - 3a^2Mr^3 + 9a^2c^2Mr^3 + 2M^2r^4 - 3Mr^5)}{\rho ^6},\\ R_{ctct}= & {} \frac{- 9a^4c^2Mr + 6a^4c^4Mr + 6a^2c^2M^2r^2 + 3a^2Mr^3 - 5a^2c^2Mr^3 - 2M^2r^4 + Mr^5}{(1 - c^2)\rho ^6},\\ R_{\varphi t \varphi t}= & {} -\frac{(1 - c^2)\Delta (3a^2c^2Mr - M r^3)}{\rho ^6}. \end{aligned}$$