BMS Charges Without Supertranslation Ambiguity

Abstract

The asymptotic symmetry of an isolated gravitating system, or the Bondi–Metzner–Sachs (BMS) group, contains an infinite-dimensional subgroup of supertranslations. Despite decades of study, the difficulties with the “supertranslation ambiguity” persisted in making sense of fundamental notions such as the angular momentum carried away by gravitational radiation. The issues of angular momentum and center of mass were resolved by the authors recently. In this paper, we address the issues for conserved charges with respect to both the classical BMS algebra and the extended BMS algebra. In particular, supertranslation ambiguity of the classical charge for the BMS algebra, as well as the extended BMS algebra, is completely identified. We then propose a new invariant charge by adding correction terms to the classical charge. With the presence of these correction terms, the new invariant charge is then shown to be free from any supertranslation ambiguity. Finally, we prove that both the classical and invariant charges for the extended BMS algebra are invariant under the boost transformations.

Appendices

Appendix A. Integral Formulae for Extended Conformal Killing Fields

We derive integral lemmas for extended gconformal Killing fields $Y^A \frac{\partial }{\partial x^A}$ defined on an open subset ${\mathscr {U}} \subset S^2$, see Definition 5.1. In these integral formulae, we impose assumptions so that integrations by parts are valid. This happens, for example, if the extended conformal Killing field is integrating against tensors with compact support in ${\mathscr {U}}$. If $Y^A \frac{\partial }{\partial x^A}$ is a global conformal Killing field, namely ${\mathscr {U}}=S^2$, those assumptions hold automatically and the formulae are applicable in Part 1.

Lemma A.1

Let $Y^A \frac{\partial }{\partial x^A}$ be an extended conformal Killing field defined on an open subset ${\mathscr {U}} \subset S^2$. If f is a function with compact support in ${\mathscr {U}}$, then we have

$$\begin{aligned} \begin{aligned}&\int _{\mathscr {U}} Y^A \left( -F_{AB}\nabla _{D}N^{BD} +N_{AB}\nabla _{D}F^{BD}-\frac{1}{2} \epsilon _{AB} \nabla ^B (\epsilon ^{PQ} F_P^{\,\,\, E}N_{EQ} ) \right) \\&\quad = \int _{\mathscr {U}} (-\nabla _AY^A f + 2 Y^A \nabla _A f )(\nabla _B\nabla _D N^{BD}) -\int _{\mathscr {U}} f\epsilon ^{PQ} N_{EQ} \nabla ^E\nabla _P(\epsilon _{CD}\nabla ^C Y^D) \end{aligned} \end{aligned}$$

(A.1)

where $F_{AB}=2\nabla _A \nabla _B f-\Delta f\sigma _{AB}$.

Proof

We denote the function $\alpha = \epsilon _{AB}\nabla ^A Y^B$ in the proof. We write the left-hand side as $(1)+(2)+(3)$. Integrating by parts, we get

$$\begin{aligned} (1)&= \int _{\mathscr {U}} -Y^A (2\nabla _A\nabla _B f -\Delta f\sigma _{AB}) \nabla _D N^{BD} \\&= \int _{\mathscr {U}} 2 \nabla ^B Y^A \nabla _A f \nabla ^D N_{BD} + 2 Y^A \nabla _A f \nabla ^B\nabla ^D N_{BD} + \Delta f Y^A \nabla ^B N_{AB},\\ (2)&= \int _{\mathscr {U}} Y^A N_{AB}\nabla ^B (\Delta +2)f\\&= \int _{\mathscr {U}} -\nabla ^B Y^A N_{AB}(\Delta +2)f -Y^A \nabla ^B N_{AB} (\Delta +2)f\\&= \int _{\mathscr {U}} -Y^A \nabla ^B N_{AB}(\Delta +2)f \end{aligned}$$

and hence

$$\begin{aligned} (1)+(2) = \int _{\mathscr {U}} 2\nabla ^B Y^A \nabla _A f \nabla ^D N_{BD} + 2 Y^A \nabla _A f \nabla ^B\nabla ^D N_{BD} - 2Y^A f \nabla ^B N_{AB}. \end{aligned}$$

We simplify the first integral

$$\begin{aligned}&\int _{\mathscr {U}} 2\nabla ^B Y^A \nabla _A f \nabla ^D N_{BD} \\&\quad = \int _{\mathscr {U}} \nabla _C Y^C \nabla ^B f \nabla ^D N_{BD} + \alpha \epsilon ^{BA} \nabla _A f \nabla ^D N_{BD}\\&\quad = \int _{\mathscr {U}} -\nabla _B\nabla _C Y^C f \nabla _D N^{BD} -\nabla _C Y^C f \nabla ^B\nabla ^D N_{BD} + \alpha \epsilon ^{BA} \nabla _A f \nabla ^D N_{BD}\\&\quad = \int _{\mathscr {U}} 2Y^B f \nabla ^D N_{BD} -\epsilon _{BC}\nabla ^C \alpha f \nabla _D N^{BD} -\nabla _C Y^C f \nabla ^B\nabla ^D N_{BD} + \alpha \epsilon ^{BA} \nabla _A f \nabla ^D N_{BD}. \end{aligned}$$

In the last equality, we used the identity

$$\begin{aligned} \nabla _B \nabla _C Y^C = -2Y_B +\epsilon _{BC}\nabla ^C \alpha , \end{aligned}$$

(A.2)

which is obtained from the computation

$$\begin{aligned} \nabla _B \nabla _C Y^C = \nabla _C \nabla _B Y^C - Y_B = \nabla ^C \left( \frac{1}{2} \nabla _D Y^D \sigma _{BD} + \frac{1}{2}\alpha \epsilon _{BC} \right) - Y_B. \end{aligned}$$

We also note the identity

$$\begin{aligned} 2 \nabla _A\nabla _B (\nabla _C Y^C) - \Delta (\nabla _C Y^C)\sigma _{AB} = \epsilon _{AC}\nabla _B \nabla ^C (\epsilon _{DE}\nabla ^D Y^E) + \epsilon _{BC}\nabla _A \nabla ^C (\epsilon _{DE}\nabla ^D Y^E), \end{aligned}$$

(A.3)

which is obtained by differentiating (A.2) and using the identity $(\Delta +2)\nabla _C Y^C =0$.

In summary, we have

$$\begin{aligned} (1)+(2)= & {} \int _{\mathscr {U}} (-\nabla _AY^A f + 2 Y^A \nabla _A f )(\nabla _B\nabla _D N^{BD})\\&-\epsilon _{BC}\nabla ^C \alpha f \nabla _D N^{BD}+ \alpha \epsilon ^{BA} \nabla _A f \nabla ^D N_{BD}. \end{aligned}$$

Finally, we have

$$\begin{aligned} (3)&= \int _{\mathscr {U}} -\frac{1}{2}\alpha \epsilon ^{PQ} (2 \nabla _P \nabla ^E f - \Delta f \delta _P^E) N_{EQ}\\&= \int _{\mathscr {U}} \nabla _P \alpha \epsilon ^{PQ} \nabla ^E f N_{EQ} + \alpha \epsilon ^{PQ}\nabla ^E f \nabla _P N_{EQ}\\&= \int _{\mathscr {U}} -\nabla ^E \nabla _P \alpha \epsilon ^{PQ} f N_{EQ} -\nabla _P \alpha \epsilon ^{PQ} f\nabla ^E N_{EQ} + \alpha \nabla ^E f \epsilon _{EP} \nabla _Q N^{EQ} \end{aligned}$$

where we used the identity $\epsilon ^{PQ}\nabla _P N_{EQ} = \epsilon ^{EP}\nabla _Q N^{EQ}$ [10, (2.13)] in the last equality. Putting these together, the assertion follows. $\square$

Lemma A.2

Let $\alpha$ be a function defined on an open subset ${\mathscr {U}} \subset S^2$. Consider two functions u and v defined on ${\mathscr {U}}$ such that one of them has compact support. Denoting $u_{AB} = \nabla _A\nabla _B u$ and $v_{AB} = \nabla _A\nabla _B v$, we have

$$\begin{aligned}&\int _{\mathscr {U}} \Big [ u(v_{AB} -\frac{1}{2} \Delta v \sigma _{AB}) - v(u_{AB} -\frac{1}{2} \Delta u \sigma _{AB}) \Big ] \nabla ^A\nabla ^B \alpha \nonumber \\&\quad =\frac{1}{2}\int _{\mathscr {U}} (u\Delta v-v\Delta u) (\Delta +2) \alpha . \end{aligned}$$

(A.4)

and

$$\begin{aligned} \begin{aligned}&\int _{\mathscr {U}} \Big [ u({\epsilon _A}^D v_{DB}+{\epsilon _B}^D v_{DA} ) - v({\epsilon _A}^D u_{DB}+{\epsilon _B}^D u_{DA} ) \Big ] \nabla ^A\nabla ^B \alpha \\&\quad =\int _{\mathscr {U}} \epsilon _{AD}\left( u \nabla ^A v - v \nabla ^A u\right) \nabla ^D (\Delta +2) \alpha .\end{aligned} \end{aligned}$$

(A.5)

In particular, applying the above formulae to the divergence of an extended conformal Killing field $Y^A \frac{\partial }{\partial x^A}$ defined on ${\mathscr {U}}$, we have

$$\begin{aligned}&\int _{\mathscr {U}} \Big [ u(v_{AB} -\frac{1}{2} \Delta v \sigma _{AB}) - v(u_{AB} -\frac{1}{2} \Delta u \sigma _{AB}) \Big ] \nabla ^A\nabla ^B (\nabla _CY^C)=0 .\\&\int _{\mathscr {U}} \Big [ u({\epsilon _A}^D v_{DB}+{\epsilon _B}^D v_{DA} ) - v({\epsilon _A}^D u_{DB}+{\epsilon _B}^D u_{DA} ) \Big ] \nabla ^A\nabla ^B (\nabla _CY^C)=0 \end{aligned}$$

if either u or v is supported in ${\mathscr {U}}$.

Proof

We write $u_A = \nabla _A u$ in the proof. We compute

$$\begin{aligned} \begin{aligned}&\int _{\mathscr {U}}(uv_{AB} - vu_{AB} ) \nabla ^A\nabla ^B \alpha \\&\quad = \int _{\mathscr {U}} \nabla _B (uv_{A} - vu_{A} ) \nabla ^A\nabla ^B \alpha \\&\quad = -\int _{\mathscr {U}} (uv_{A} - vu_{A} )\nabla _B\nabla ^A\nabla ^B \alpha \\&\quad =-\int _{\mathscr {U}} (uv_{A} - vu_{A} )\big [ \nabla ^A\nabla _B\nabla ^B(\nabla _CY^C) + \nabla ^A (\nabla _CY^C) \Big ]\\&\quad =-\int _{\mathscr {U}} (uv_{A} - vu_{A} ) \nabla ^A(\Delta +1)\alpha \\&\quad =\int _{\mathscr {U}} ( u \Delta v - v \Delta u)(\Delta +1)\alpha \\ \end{aligned} \end{aligned}$$

(A.4) follows from rearranging the terms. For (A.5), note that we can replace $\nabla ^A\nabla ^B \alpha$ by the symmetric traceless 2-tensor $\nabla ^A\nabla ^B \alpha - \frac{1}{2}\Delta \alpha \sigma ^{AB}$. Recalling Proposition 2.4 of [10] that $\epsilon _{DB} \nabla ^D C^{BA} = \epsilon ^{AD}\nabla ^B C_{BD}$ for any symmetric traceless 2-tensor, we integrate by parts to get

$$\begin{aligned}&\int _{\mathscr {U}} u ( {\epsilon _A}^D v_{DB}+{\epsilon _B}^D v_{DA} ) \nabla ^A\nabla ^B \alpha \\&\quad = \int _{\mathscr {U}} -\left( u_D {\epsilon _A}^D v_B + u_A {\epsilon _B}^D v_D \right) \left( \nabla ^A\nabla ^B \alpha - \frac{1}{2}\Delta \alpha \sigma ^{AB} \right) \\&\qquad + \int _{\mathscr {U}} -u v_B \left( -\frac{1}{2}{\epsilon _B}^D \nabla _D (\Delta +2 )\alpha \right) -u{\epsilon _B}^D v_D \cdot \frac{1}{2}\nabla ^B (\Delta +2)\alpha . \end{aligned}$$

Interchanging u and v and subtraction yield (A.5). The second claims follows from $(\Delta +2)(\nabla _CY^C)=0$ for an extended BMS field. $\square$

The following lemma generalizes Lemma 2.3 of [10].

Lemma A.3

Let $Y^A \frac{\partial }{\partial x^A}$ be an extended conformal Killing field defined on an open subset ${\mathscr {U}} \subset S^2$. If u is a function with compact support in ${\mathscr {U}}$, then we have

$$\begin{aligned} \begin{aligned}&\int _{\mathscr {U}} Y^1 [2\nabla _A\nabla _Bu \nabla ^A\nabla ^B u- (\Delta u)^2] .\\&\quad = \int _{\mathscr {U}} Y^1[(\Delta +2)u]^2 + 2 \int _{\mathscr {U}} u (\nabla ^A\nabla ^B Y^1 ) (\nabla _A\nabla _Bu - \frac{1}{2} \Delta u \sigma _{AB}) \end{aligned} \end{aligned}$$

where $2Y^1=\nabla _CY^C$

Proof

We use the following formulae in the derivation

$$\begin{aligned} \begin{aligned} \Delta |\nabla u|^2&=2|\nabla ^2u|^2+2\nabla u\cdot \nabla (\Delta +1)u\\ \Delta (u^2)&=2|\nabla u|^2+2 u\Delta u\\ \Delta (u\Delta u)&=(\Delta u)^2+2\nabla u\cdot \nabla (\Delta u)+u \Delta ^2 u. \end{aligned} \end{aligned}$$

Integrating by parts twice gives

$$\begin{aligned} \int _{\mathscr {U}} Y^1\nabla _A\nabla _Bu \nabla ^A\nabla ^B u=\int _{\mathscr {U}} u \nabla ^A\nabla ^B(Y^1 \nabla _A\nabla _Bu) \end{aligned}$$

We compute

$$\begin{aligned} \begin{aligned}&\nabla ^A\nabla ^B(Y^1 \nabla _A\nabla _Bu)\\&\quad =(\nabla ^A\nabla ^B Y^1) \nabla _A\nabla _Bu+2\nabla ^B Y^1 \nabla ^A \nabla _A\nabla _Bu+Y^1 \nabla ^A\nabla ^B \nabla _A\nabla _Bu\\&\quad =-Y^1\Delta u+2\nabla ^BY^1\nabla _B (\Delta +1)u+Y^1\Delta (\Delta +1)u + (\nabla ^A\nabla ^B Y^1- \frac{1}{2}\Delta Y^1 \sigma _{AB} ) \nabla _A\nabla _Bu \\&\quad =Y^1 \Delta ^2 u+2\nabla ^B Y^1 \nabla _B (\Delta +1)u + (\nabla ^A\nabla ^B Y^1- \frac{1}{2}\Delta Y^1 \sigma _{AB} ) \nabla _A\nabla _Bu \end{aligned}, \end{aligned}$$

where we use $\nabla ^A \nabla _A\nabla _Bu=\nabla _B (\Delta +1)u$ and $\Delta Y^1 = -2 Y^1$ in the second equality.

On the other hand, we have the identity:

$$\begin{aligned} 2\nabla ^B u\nabla _Bv=\Delta (uv)- u\Delta v-v\Delta u \end{aligned}$$

and thus

$$\begin{aligned} 2\nabla ^BY^1\nabla _B (\Delta +1)u=\Delta (Y^1 (\Delta +1)u)-Y^1 \Delta (\Delta +1)u+2Y^1 (\Delta +1)u. \end{aligned}$$

Putting all together gives:

$$\begin{aligned} \begin{aligned}&\int _{\mathscr {U}} Y^1 \nabla _A\nabla _Bu \nabla ^A\nabla ^B u\\&\quad =\int _{\mathscr {U}} Y^1\Delta ^2 u+\int _{\mathscr {U}} u [\Delta (Y^1 (\Delta +1)u)-Y^1 \Delta (\Delta +1)u+2Y^1 (\Delta +1)u]\\&\qquad + \int _{\mathscr {U}} u (\nabla ^A\nabla ^B Y^1- \frac{1}{2}\Delta Y^1 \sigma _{AB} ) \nabla _A\nabla _Bu\\&\quad =\int _{\mathscr {U}} Y^1 [(\Delta u)^2+2u\Delta u+2u^2] + \int _{\mathscr {U}} u (\nabla ^A\nabla ^B Y^1- \frac{1}{2}\Delta Y^1 \sigma _{AB} ) \nabla _A\nabla _Bu \end{aligned}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&\int _{\mathscr {U}} Y^1 [2\nabla _A\nabla _Bu \nabla ^A\nabla ^B u- (\Delta u)^2] .\\&\quad = \int _{\mathscr {U}} Y^1[(\Delta +2)u]^2 + 2 \int _{\mathscr {U}} u (\nabla ^A\nabla ^B Y^1- \frac{1}{2}\Delta Y^1 \sigma _{AB} ) \nabla _A\nabla _Bu\\&\quad = \int _{\mathscr {U}} Y^1[(\Delta +2)u]^2 + 2 \int _{\mathscr {U}} u (\nabla ^A\nabla ^B Y^1 ) (\nabla _A\nabla _Bu - \frac{1}{2}\Delta u \sigma _{AB}) \end{aligned} \end{aligned}$$

$\square$

The next lemma, used in all invariance proofs in non-radiative spacetimes, generalizes Lemma B.1 and B.2 in [10]. It also appears in the derivation from (B.8) to (B.9) in [6].

Lemma A.4

Let $Y^A \frac{\partial }{\partial x^A}$ be an extended conformal Killing vector field defined on an open subset ${\mathscr {U}} \subset S^2$. Let f be a function on $S^2$ and $C_{AB}$ be a symmetric traceless 2-tensor on $S^2$. If either f or $C_{AB}$ has compact support in ${\mathscr {U}}$, then

$$\begin{aligned}&\frac{1}{4}\int _{\mathscr {U}} Y^A \left[ \! C_{AB}\nabla _D F^{BD}+F_{AB} \nabla _D C^{BD}+\frac{1}{2} \nabla _A ( C_{BD} F^{BD})-f \nabla ^B P_{BA} -3P_{BA} \nabla ^B f \!\right] \\&\quad = -\frac{1}{4} \int _{\mathscr {U}} f \nabla ^A\nabla ^B(\nabla _C Y^C) C_{AB} \end{aligned}$$

where $F_{AB} = 2\nabla _A \nabla _B f - \Delta f \sigma _{AB}$ and $P_{BA} = \nabla _B \nabla ^E C_{EA} - \nabla _A \nabla ^E C_{EB}$.

Remark A.5

If $Y^A \frac{\partial }{\partial x^A}$ is defined globally on $S^2$, then we have

$$\begin{aligned} \nabla ^A\nabla ^B(\nabla _C Y^C) C_{AB} = 0. \end{aligned}$$

Proof

Integrating by part the last two terms of the first line, we have

$$\begin{aligned} \begin{aligned}&\int _{{\mathscr {U}}}Y^A(-f \nabla ^B P_{BA} -3P_{BA} \nabla ^B f )\\&\quad = \int _{{\mathscr {U}}}-2 Y^A (\nabla _B\nabla ^D C_{DA} - \nabla _A\nabla ^D C_{DB})\nabla ^B f +2f (\nabla ^B Y^A \\&\qquad - \frac{1}{2} \nabla _EY^E \sigma ^{AB})\nabla _B\nabla ^D C_{DA} \\&\quad = \int _{{\mathscr {U}}} 2 \nabla _B Y^A\nabla ^D C_{DA}\nabla ^B f + 2 Y^A\nabla ^D C_{DA} \Delta f \\&\qquad - 2 \nabla _A Y^A\nabla ^D C_{DB}\nabla ^B f-2 Y^A\nabla ^D C_{DB} \nabla _A\nabla ^B f\\&\qquad +\int _{{\mathscr {U}}} 2fY^A\nabla ^D C_{DA}-2 \nabla ^B Y^A \nabla _B f\nabla ^D C_{DA}-f \nabla _EY^E\nabla ^A\nabla ^B C_{AB} \\&\quad = \int _{{\mathscr {U}}} - 2 Y^A \nabla ^D C_{DB} (\nabla _A\nabla ^B f - \frac{1}{2}\Delta f \sigma _{AB}) + Y^A \nabla ^D C_{AD} (\Delta + 2)f \\&\qquad -\int _{{\mathscr {U}}}2 \nabla _A Y^A\nabla ^D C_{DB}\nabla ^B f+f \nabla _EY^E\nabla ^A\nabla ^B C_{AB}, \end{aligned} \end{aligned}$$

where in the first equality we anti-symmetrize $\nabla ^B Y^A$ into $\nabla ^B Y^A - \frac{1}{2} \nabla _E Y^E \sigma ^{AB}$ by (5.2).

After simplifying the last two terms

$$\begin{aligned} \begin{aligned}&\int _{{\mathscr {U}}}2 \nabla _A Y^A\nabla ^D C_{DB}\nabla ^B f+f \nabla _EY^E\nabla ^A\nabla ^B C_{AB}\\&\quad =\int _{{\mathscr {U}}}-2 \nabla ^D\nabla _A Y^A C_{DB}\nabla ^B f -2\nabla _AY^A C_{DB} \nabla ^D\nabla ^B f \\&\qquad + \int _{{\mathscr {U}}} \nabla ^A \nabla ^B f \nabla _EY^E C_{AB} + f \nabla ^A\nabla ^B\nabla _EY^E C_{AB} + 2 \nabla ^Bf \nabla ^A (\nabla _EY^E) C_{AB} \\&\quad = \int _{{\mathscr {U}}} f \nabla ^A\nabla ^B\nabla _EY^E C_{AB} - \nabla _AY^A C_{DB} \nabla ^D\nabla ^B f\\&\quad = \int _{{\mathscr {U}}} f \nabla ^A\nabla ^B\nabla _EY^E C_{AB} - \frac{1}{2}\nabla _AY^A C_{DB} F^{DB}, \end{aligned} \end{aligned}$$

we get

$$\begin{aligned}&\frac{1}{4}\int _{{\mathscr {U}}} Y^A \left[ \! C_{AB}\nabla _D F^{BD}{+}F_{AB} \nabla _D C^{BD}{+}\frac{1}{2} \nabla _A (C_{BD} F^{BD})-f \nabla ^B P_{BA} -3P_{BA} \nabla ^B f\!\right] \\&\quad = -\frac{1}{4} \int _{{\mathscr {U}}} f \nabla ^A\nabla ^B(\nabla _C Y^C) C_{AB}. \end{aligned}$$

$\square$

Appendix B. Explicit Forms of Extended BMS Fields

An extended BMS field corresponds to a singular solution of (5.2) which cannot be integrated to a conformal diffeomorphism of $S^2$. In the following, we discuss these solutions. First observe that the equation (5.2) is conformally invariant in the sense that $Y^A$ is a solution of (5.2) with respect to $\sigma$ if and on if $Y^A$ is a solution of (5.2) with respect to a metric that is conformal to $\sigma$. Consider the stereographic projection $\rho$ from the complement of the north pole (0, 0, 1) of ${S}^2\subset {\mathbb {R}}^3$ to ${\mathbb {R}}^2={\mathbb {C}}$:

$$\begin{aligned} \rho : S^2-\{(0, 0, 1)\}\rightarrow {\mathbb {R}}^2={\mathbb {C}}. \end{aligned}$$

The pull-back of the flat metric $\mathring{\sigma }=|dz|^2= (dx)^2+(dy)^2$ on ${\mathbb {R}}^2={\mathbb {C}}$ through $\rho$ is conformal to the standard round metric $\sigma$ on $S^2$. With respect to the flat metric $\bar{\sigma }=\rho ^*\mathring{\sigma }$, (5.2) is exactly the Cauchy–Riemann equation, i.e $Y^1\partial _x+Y^2\partial _y$ satisfies (5.2) if and only if $Y^1+iY^2$ satisfies the Cauchy–Riemann equation $\partial _{{\bar{z}}}(Y^1+iY^2)=0$. A complete basis for the extended conformal Killing fields can be found in [21] in terms of $\ell =1$ spherical harmonics.

$$\begin{aligned} l_m=-z^{m+1} \partial _z, {\bar{l}}_m=-{\bar{z}}^{m+1} \partial _{{\bar{z}}}, m \in {\mathbb {Z}}. \end{aligned}$$

In particular, $\partial _z$ and $z^{m+1}\partial _z$ are conformal Killing fields with respect to $\bar{\sigma }$. We also recall that the function z on ${\mathbb {C}}$ satisfies $\bar{\Delta }z=0$ and $\bar{\nabla }_A z\bar{\nabla }^Az=0$ with respect to $\bar{\sigma }$.

Let Z be the pull-back of z through $\rho$, and $\mathring{Y}^A$ be the pull back of $(\partial _z)^A$ through $\rho ^{-1}$. Z and $\mathring{Y}^A$ satisfy

$$\begin{aligned} \Delta Z=0, \nabla _AZ\nabla ^A Z=0, \nabla ^A \mathring{Y}^B+\nabla ^B \mathring{Y}^A-\nabla _C\mathring{Y}^C \sigma ^{AB}=0 \end{aligned}$$

by the conformal invariance of these equations.

Explicitly Z, ${\bar{Z}}$, and $\mathring{Y}^A$ are given by

$$\begin{aligned} Z=\frac{{\tilde{X}}^1+i{\tilde{X}}^2}{1-{\tilde{X}}^3}, {\bar{Z}}=\frac{{\tilde{X}}^1-i{\tilde{X}}^2}{1-{\tilde{X}}^3} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \mathring{Y}^A&=(\epsilon ^{CD} \nabla _C Z\nabla _D {\bar{Z}})^{-1}\epsilon ^{AB} \nabla _B {\bar{Z}} \end{aligned}. \end{aligned}$$

All are defined and smooth outside the north pole (0, 0, 1). A straightforward calculation shows that

$$\begin{aligned} \nabla _A \mathring{Y}^A=-({\tilde{X}}^1-i{\tilde{X}}^2) \text { and } \nabla _A Z\nabla ^A{\tilde{X}}^3=Z. \end{aligned}$$

(B.1)

Lemma B.1

The divergence $\nabla _A Y^A$ of the conformal Killing field $Y^A=Z^{m+1} \mathring{Y}^A$ satisfies

$$\begin{aligned} (\Delta +2)\nabla _A Y^A=0 \end{aligned}$$

on $S^2-\{(0, 0, 1)\}$.

Proof

We compute

$$\begin{aligned} \begin{aligned}\nabla _A Y^A&=\nabla _A (Z^{m+1} \mathring{Y}^A)\\&=(m+1) Z^m (\nabla _A Z) \mathring{Y}^A+Z^{m+1} \nabla _A\mathring{Y}^A \\&=(m+1) Z^m-Z^{m+1} ({\tilde{X}}^1-i{\tilde{X}}^2)\\&=Z^m(m-{\tilde{X}}^3),\end{aligned} \end{aligned}$$

where we use the first equation in (B.1).

Furthermore,

$$\begin{aligned} \begin{aligned} \Delta (\nabla _A Y^A)&=(\Delta Z^m)(m-{\tilde{X}}^3)-2m Z^{m-1}\nabla _A Z \nabla ^A {\tilde{X}}^3-Z^m \Delta {\tilde{X}}^3\\&=-2 Z^m(m-{\tilde{X}}^3), \end{aligned} \end{aligned}$$

where we use the second equation in (B.1). $\square$

Appendix C. Conformal Transformation on $S^2$

Let $\sigma$ be the induced metric of unit sphere $S^2 \subset {\mathbb {R}}^3$ Let $g: S^2 \rightarrow S^2$ be a conformal map and ${\widehat{\sigma }} = g^*\sigma$. It is well-known that g is a linear fractional transformation and a direct computation yields that ${\widehat{\sigma }} = K^2 {\bar{\sigma }}$. In addition, both ${\widehat{\sigma }}$ and ${\bar{\sigma }}$ have constant curvature 1. Moreover,

$$\begin{aligned} K(x) = \frac{1}{\alpha _0 + \alpha _i x^i} >0 \end{aligned}$$

where $(\alpha _0,\alpha _i)$ is a unit timelike vector and $x^i,i=1,2,3$ forms an orthonormal basis of first eigenfunctions on $(S^2,{\bar{\sigma }})$.

We denote by ${\bar{\nabla }},{\bar{\Delta }}$ (${\widehat{\nabla }}, {\widehat{\Delta }}$ respectively) the covariant derivative and Laplacian with respect to ${\bar{\sigma }}$ (${\widehat{\sigma }},$ respectively). Taking Hessian and Laplacian of $K^{-1}$, we get

Lemma C.1

$$\begin{aligned}&K^{-2} {\bar{\nabla }}_a {\bar{\nabla }}_b K - 2K^{-3} {\bar{\nabla }}_a K {\bar{\nabla }}_b K = \alpha _i x^i {\bar{\sigma }}_{ab} \end{aligned}$$

(C.1)

$$\begin{aligned}&\frac{1}{2} K^{-2} {\bar{\Delta }} K - K^{-3}|{\bar{\nabla }} K|^2 = \alpha _i x^i \end{aligned}$$

(C.2)

The next lemma compares the covariant derivatives ${\bar{\nabla }}$ and ${\widehat{\nabla }}.$ The proof is through direct computation and is left to the readers.

Lemma C.2

• (1) The Christoffel symbols of ${\widehat{\sigma }}$ and ${\bar{\sigma }}$ are related by

  $$\begin{aligned} \widehat{\Gamma }_{ab}^c={\bar{\Gamma }}_{ab}^c + \frac{1}{2} K^{-2}\left( (\partial _b K^2) \delta _a^c+(\partial _a K^2) \delta _b^c-(\partial _d K^2) {\bar{\sigma }}_{ab} {\bar{\sigma }}^{cd}\right) . \end{aligned}$$

• (2) The covariant derivatives of a covector $X_a$ are related by

  $$\begin{aligned} \begin{aligned} {\widehat{\nabla }}_b X_a&={\bar{\nabla }}_b X_a + K^{-1} (\partial _b K)X_a + K^{-1} (\partial _aK) X_b - K^{-1}(\nabla ^cK)X_c \,{\bar{\sigma }}_{ab}\\ K^2 \widehat{\sigma }^{ab} \widehat{\nabla }_b X_a&={{\bar{\sigma }}}^{ab} {{\bar{\nabla }}}_b X_a. \end{aligned} \end{aligned}$$

  I n particular, for a function f, we have

  $$\begin{aligned} K^2\widehat{\Delta } f={\bar{\Delta }} f \end{aligned}$$

  (C.3)

  and

  $$\begin{aligned} {\widehat{\nabla }}_a {\widehat{\nabla }}_b (Kf) - \frac{1}{2} \widehat{\Delta } (Kf )\widehat{\sigma }_{ab} =K\left( {\bar{\nabla }}_a {\bar{\nabla }}_b f -\frac{1}{2} {\bar{\Delta }} f {\bar{\sigma }}_{ab}\right) \end{aligned}$$

  (C.4)

• (3) If $C_{ab}$ is a traceless symmetric 2-tensor for ${\bar{\sigma }}$ (hence also traceless for ${\widehat{\sigma }}$), then

  $$\begin{aligned} {\widehat{\nabla }}^b (KC_{ab}) = K^{-1}{\bar{\nabla }}^b C_{ab} + K^{-2} {\bar{\nabla }}^b K C_{ab} = K^{-2} {\bar{\nabla }}^b (KC_{ab}) \end{aligned}$$

  (C.5)

  and

  $$\begin{aligned} {\widehat{\nabla }}^a {\widehat{\nabla }}^b (KC_{ab}) = K^{-3} {\bar{\nabla }}^a {\bar{\nabla }}^b C_{ab}. \end{aligned}$$

  (C.6)