Mass, center of mass and isoperimetry in asymptotically flat 3-manifolds

Abstract

We revisit the interplay between the mass, the center of mass and the large scale behavior of certain isoperimetric quotients in the setting of asymptotically flat 3-manifolds (both without and with a non-compact boundary). In the boundaryless case, we first check that the isoperimetric deficits involving the total mean curvature recover the ADM mass in the asymptotic limit, thus extending a classical result due to G. Huisken. Next, under a Schwarzschild asymptotics and assuming that the mass is positive we indicate how the implicit function method pioneered by R. Ye and refined by L.-H. Huang may be adapted to establish the existence of a foliation of a neighborhood of infinity satisfying the corresponding curvature conditions. Recovering the mass as the asymptotic limit of the corresponding relative isoperimetric deficit also holds true in the presence of a non-compact boundary, where we additionally obtain, again under a Schwarzschild asymptotics, a foliation at infinity by free boundary constant mean curvature hemispheres, which are shown to be the unique relative isoperimetric surfaces for all sufficiently large enclosed volume, thus extending to this setting a celebrated result by M. Eichmair and J. Metzger. Also, in each case treated here we relate the geometric center of the foliation to the center of mass of the manifold as defined by Hamiltonian methods.

Appendices

Appendix A: The large scale isoperimetric deficits and the mass: the proofs of Theorems 4 and 7

The arguments to prove Theorems 4 and 7 are simple variations on the computation appearing in [39, Section 2], where a proof of Theorem 2 appears. This justifies the inclusion of a somewhat detailed account of their calculation in what follows.

If (M, g) is asymptotically flat as in Definition 1, we first observe that, since $\partial r/\partial x_i=x_i/r$, we have

$$\begin{aligned} \nabla r=g^{ij}\frac{x_i}{r}\frac{\partial }{\partial x_j}. \end{aligned}$$

(A1)

and hence

$$\begin{aligned} \vert \nabla r\vert ^2=g^{ij}\frac{x_ix_j}{r^2}=1-e_{ij}\frac{x_ix_j}{r^2}+O(r^{-2\tau }). \end{aligned}$$

(A2)

Also, if $\nu$ is the outward unit normal vector field to the coordinate 2-sphere $S^2_r$ then

$$\begin{aligned} \nu =\frac{x}{r}+O(r^{-\tau }). \end{aligned}$$

(A3)

Let $dS^{2,\delta }_r=r^2dS^{2,\delta }_1$ be the area element of the Euclidean sphere of radius r. It follows that the area element of the corresponding coordinate sphere $S^2_r$ expands as

$$\begin{aligned} dS_r^2=\left( 1+\frac{1}{2}h^{ij}e_{ij}+O(r^{-2\tau })\right) dS_r^{2,\delta }, \end{aligned}$$

(A4)

where

$$\begin{aligned} h_{ij}=g_{ij}-\nu _i\nu _j=\delta _{ij}-\frac{x_ix_j}{r^2}+O(r^{-\tau }) \end{aligned}$$

(A5)

is the induced metric (extended to vanish in the radial direction). Thus, the area of $S^{2}_r$ is

$$\begin{aligned} A(r)=4\pi r^2+\frac{1}{2}\int _{S_r^2}h^{ij}e_{ij}dS_r^{2,\delta }+O(r^{2-2\tau }). \end{aligned}$$

(A6)

From this we obtain

$$\begin{aligned} \frac{d}{dr}A(r)= & {} 8\pi r+\frac{1}{2}\int _{S_r^2}h^{ij}\frac{x_k}{r}e_{ij,k}dS_r^{2,\delta }+\frac{1}{r}\int _{S^2_r}h^{ij}e_{ij}dS_r^{2,\delta }+O(r^{1-2\tau }), \end{aligned}$$

where the comma means partial differentiation. Using (A5) we get

$$\begin{aligned} \frac{d}{dr}A(r)= & {} 8\pi r + \frac{1}{2}\int _{S_r^2}e_{ii,k}\frac{x_k}{r}dS_r^{2,\delta }-\frac{1}{2}\int _{S_r^2}e_{ij,k}\frac{x_ix_jx_k}{r^3}dS_r^{2,\delta }\\{} & {} +\frac{1}{r}\int _{S^2_r}h^{ij}e_{ij}dS_r^{2,\delta }+O(r^{1-2\tau }). \end{aligned}$$

We now work out the third term in the right-hand side. We first note that

$$\begin{aligned} \frac{\partial }{\partial x_i}\frac{x_j}{r}=\frac{\delta _{ij}}{r}-\frac{x_ix_j}{r^3}. \end{aligned}$$

(A7)

We then compute:

$$\begin{aligned} \int _{S_r^2}\frac{\partial }{\partial x_k}\left( e_{ij}\frac{x_j}{r}\right) \frac{x_ix_k}{r^2}dS_r^{2,\delta }= & {} \int _{S_r^2}e_{ij,k}\frac{x_ix_jx_k}{r^3}dS_r^{2,\delta }\\{} & {} \quad +\int _{S_r^0}e_{ij}\left( \frac{\delta _{jk}}{r}-\frac{x_jx_k}{r^3}\right) \frac{x_ix_k}{r^2}dS_r^{2,\delta }\\= & {} \int _{S_r^2}e_{ij,k}\frac{x_ix_jx_k}{r^3}dS_r^{2,\delta }, \end{aligned}$$

so we have

$$\begin{aligned} \int _{S_r^2}e_{ij,k}\frac{x_ix_jx_k}{r^3}dS_r^{2,\delta }= & {} \int _{S_r^2}\frac{\partial }{\partial x_k}\left( e_{ij}\frac{x_j}{r}\right) \frac{x_ix_k}{r^2}dS_r^{2,\delta }\\= & {} \underbrace{\int _{S_r^2}\frac{\partial }{\partial x_i}\left( e_{ij}\frac{x_j}{r}\right) dS_r^{2,\delta }}_{(I)}\\{} & {} \underbrace{- \int _{S^2_r}\left( \delta _{ik}-\frac{x_ix_k}{r^2}\right) \frac{\partial }{\partial x_k}\left( e_{ij}\frac{x_j}{r}\right) dS_r^{2,\delta }}_{(II)}. \end{aligned}$$

Using (A5) we have

$$\begin{aligned} (I)=\int _{S_r^2}e_{ij,i}\frac{x_j}{r}dS_r^{2,\delta }+\frac{1}{r}\int _{S_r^2}h^{ij}e_{ij}dS_r^{2,\delta }+O(r^{1-2\tau }). \end{aligned}$$

Also, integration by parts together with (A7) gives

$$\begin{aligned} (\textit{II})=-\int _{S^2_r}\frac{\partial }{\partial x_k}\left( \frac{x_ix_k}{r^2}\right) e_{ij}\frac{x_j}{r}dS_r^{2,\delta }=-2\int _{S^2_r}e_{ij}\frac{x_ix_j}{r^3}dS_r^{2,\delta }, \end{aligned}$$

(A8)

so that

$$\begin{aligned} \int _{S_r^2}e_{ij,k}\frac{x_ix_jx_k}{r^3}dS_r^{2,\delta }= & {} -2\int _{S^2_r}e_{ij}\frac{x_ix_j}{r^3} dS_r^{2,\delta }+\int _{S^2_r}e_{ij,i}\frac{x_j}{r}dS_r^{2,\delta }\\{} & {} +\frac{1}{r}\int _{S^2_r}h^{ij}e_{ij} dS_r^{2,\delta }+O(r^{1-2\tau }). \end{aligned}$$

Thus,

$$\begin{aligned} \frac{d}{dr}A(r)= & {} {{8\pi r+\frac{1}{2}\int _{S^2_r}(e_{ii,j}-e_{ij,i})\frac{x_j}{r}dS_r^{2,\delta }\nonumber }}\\{} & {} {{ +\int _{S^2_r}e_{ij}\frac{x_ix_j}{r^3}dS_r^{2,\delta }+\frac{1}{2r}\int _{S^2_r}h^{ij}e_{ij}dS_r^{2,\delta }+O(r^{1-2\tau })}}\nonumber \\= & {} 8\pi r-8\pi m+\int _{S^2_r}e_{ij}\frac{x_ix_j}{r^3}dS_r^{2,\delta }+\frac{1}{2r}\int _{S^2_r}h^{ij}e_{ij}dS_r^{2,\delta }+o(1). \end{aligned}$$

(A9)

Combining this with (A6), we get

$$\begin{aligned} \frac{d}{dr}A(r)=\frac{A(r)}{r}+4\pi r-8\pi m+\int _{S^2_r}e_{ij}\frac{x_ix_j}{r^3}dS_r^{2,\delta }+o(1). \end{aligned}$$

(A10)

We now look at the volume V(r) enclosed by $S_r^2$. By the co-area formula, (A2) and (A4),

$$\begin{aligned} \frac{1}{r}\frac{d}{dr}V(r)= & {} \frac{1}{r}\int _{S_r^2}\vert \nabla r\vert ^{-1} dS^2_r\nonumber \\= & {} \frac{A(r)}{r}+\frac{1}{2}\int _{S_r^2}e_{ij}\frac{x_jx_j}{r^3}dS_r^{2,\delta }+o(1). \end{aligned}$$

(A11)

We may now eliminate the integral term in (A10) and (A11). The result is

$$\begin{aligned} \frac{d}{dr}(rA(r))=4\pi r^2-8\pi mr+2\frac{d}{dr}V(r)+o(r). \end{aligned}$$

Integrating we obtain a formula relating the volume and area, namely,

$$\begin{aligned} V(r)=\frac{1}{2}rA(r)-\frac{2\pi }{3}r^3+2\pi m r^2+o(r^2), \end{aligned}$$

(A12)

which gives

$$\begin{aligned} J_r^{M;3,2}=r+\frac{4\pi r^2}{A(r)}\left( m-\frac{r}{3}\right) -\frac{2r}{3}\left( \frac{A(r)}{4\pi r^2}\right) ^{\frac{1}{2}}+o(1). \end{aligned}$$

On the other hand, from (A6),

$$\begin{aligned} \frac{A(r)}{4\pi r^2}=1+{\mathcal {I}}+O(r^{-2\tau }),\quad \mathcal I:=\frac{1}{8\pi r^2}\int _{S_r^2}h^{ij}e_{ij}dS_r^{2,\delta }=O(r^{-\tau }) \end{aligned}$$

so that

$$\begin{aligned} J_r^{M;3,2}= & {} r+\left( 1-{\mathcal {I}}+O(r^{-2\tau })\right) \left( m-\frac{r}{3}\right) -\frac{2r}{3}\left( 1+\frac{1}{2}{\mathcal {I}}+O(r^{-2\tau })\right) +o(1)\\= & {} m+o(1), \end{aligned}$$

which gives the proof of Theorem 2.

So far we have been following [39] closely. We now explain how a little variation yields the proof of Theorem 4. We will make use of the well-known expansion

$$\begin{aligned} H=\frac{2}{r}+O(r^{-\tau -1}). \end{aligned}$$

(A13)

Together with (A4) this gives

$$\begin{aligned} \frac{M(r)}{8\pi r}=1+{\mathcal {I}}+O(r^{-2\tau }). \end{aligned}$$

(A14)

Also, by the first variation formula for the area,

$$\begin{aligned} \frac{d}{dr}A(r)= & {} \int _{S_r^2}\left\langle \frac{\partial }{\partial r},H\nu \right\rangle dS^2_r\\{} & {} {\mathop {=}\limits ^{(A2)+(A13)}} \int _{S_r^2}\vert \nabla r\vert ^{-1}H dS^2_r\\= & {} M(r)+\int _{S^2_r}e_{ij}\frac{x_ix_j}{r^3}dS^{2,\delta }_r+O(r^{-2\tau +1}), \end{aligned}$$

and combining this with (A10) we get

$$\begin{aligned} \frac{1}{2}r^2M(r)=\frac{1}{2}r{A(r)}+2\pi r^3-4\pi mr^2+o(r^2). \end{aligned}$$

(A15)

We now use (A12) to eliminate the area term. Solving for the volume we get

$$\begin{aligned} V(r)=\frac{1}{2}r^2M(r)-\frac{8\pi }{3}r^3+6\pi mr^2+o(r^2), \end{aligned}$$

(A16)

so that, using (A14),

$$\begin{aligned} J_r^{M;3,1}= & {} \frac{2}{3}r+\frac{8\pi r}{M(r)}\left( m-\frac{4}{9}r\right) -\frac{2}{9}r\left( \frac{M(r)}{8\pi r}\right) ^2+o(1)\\= & {} \frac{2}{3}r+\left( 1-{\mathcal {I}}+O(r^{-2\tau })\right) \left( m-\frac{4}{9}r\right) -\frac{2}{9}r\left( 1+2{\mathcal {I}}+O(r^{-2\tau })\right) +o(1)\\= & {} m + o(1), \end{aligned}$$

which finishes the proof of the first equality in (11). As for the second one, note that by (A15) and (A14),

$$\begin{aligned} J_r^{M;2,1}= & {} r+\frac{4\pi r}{M(r)}\left( 2m-r\right) -\frac{1}{16\pi }M(r)+o(1)\\= & {} r+\frac{1}{2}\left( 1-{\mathcal {I}}+O(r^{-2\tau })\right) \left( 2m-r\right) -\frac{r}{2}\left( 1+{\mathcal {I}}+O(r^{-2\tau })\right) +o(1)\\= & {} m+o(1), \end{aligned}$$

which completes the proof of Theorem 4.

We now present the proof of Theorem 7. We first observe that instead of (A6) we now have

$$\begin{aligned} {\mathcal {A}} (r)=2\pi r^2+\frac{1}{2}\int _{S_{r,+}^2}h^{ij}e^+_{ij}dS_{r,+}^{2,\delta ^+}+O(r^{2-2\tau }). \end{aligned}$$

(A17)

Also, the integration by parts leading to (A8) now produces an extra term, so that (II) gets replaced by

$$\begin{aligned} (II^+)=-2\int _{S^2_{r,+}}e^+_{ij}\frac{x_ix_j}{r^3}dS_{r,+}^{2,\delta ^+}-\int _{ S^1_{r}}e^+_{kj}\frac{x_j}{r}\vartheta ^k dS^{1,\delta ^+}_r+O(r^{-2\tau +1}). \end{aligned}$$

(A18)

Thus, instead of (A10) we now have

$$\begin{aligned} \frac{d}{dr}{\mathcal {A}}(r)=\frac{{\mathcal {A}}(r)}{r}+2\pi r-8\pi \mathfrak m+\int _{S^2_{r,+}}e^+_{ij}\frac{x_ix_j}{r^3}dS_{r,+}^{2,\delta ^+}+o(1). \end{aligned}$$

(A19)

Hence, proceeding exactly as before we now get

$$\begin{aligned} {\mathcal {V}}(r)=\frac{1}{2}r{\mathcal {A}}(r)-\frac{\pi }{3}r^3+2\pi {\mathfrak {m}} r^2+o(r^2), \end{aligned}$$

(A20)

which gives

$$\begin{aligned} {\mathcal {J}}^{M;3,2}_r= & {} \frac{r}{2}+\frac{2\pi r^2}{{\mathcal {A}}(r)}\left( {\mathfrak {m}} -\frac{r}{6}\right) -\frac{r}{3}\left( \frac{{\mathcal {A}}(r)}{2\pi r^2}\right) ^{\frac{1}{2}}+o(1)\\= & {} \frac{r}{2}+\left( 1-{{\widehat{I}}}+O(r^{-2\tau })\right) \left( {\mathfrak {m}} -\frac{r}{6}\right) -\frac{r}{3}\left( \left( 1+\frac{1}{2}{{\widehat{I}}}+O(r^{-2\tau })\right) \right) +o(1)\\= & {} {\mathfrak {m}} +o(1), \end{aligned}$$

where

$$\begin{aligned} {{\widehat{I}}}=\frac{1}{4\pi r^2}\int _{S^2_{r,+}}h^{ij}e^+_{ij}dS^2_{r,+}=O(r^{-\tau }). \end{aligned}$$

This completes the proof of Theorem 7.

Appendix B: The variational setup

Here we address the variational issues needed in the bulk of the paper. Our aim is twofold. First, we review the well-known variational theory of free boundary constant mean curvature surfaces [58, 59] in a way that is convenient for our purposes. Next, we discuss the much less known variational theory of closed surfaces which are critical for the total mean curvature functional under a volume preserving constraint and develop the corresponding stability theory. We remark that the variational theory associated to curvature integrals involving elementary symmetric functions of the principal curvatures (quermassintegrals) of hypersurfaces in space forms is a well established subject; see [60] and the references therein.

We start by considering a one-parameter family of compact, embedded surfaces $t\in (-\varepsilon ,\varepsilon )\mapsto S_t$ in an arbitrary Riemannian manifold $(M^3,g)$ evolving as

$$\begin{aligned} \frac{\partial x_t }{\partial t}=Y_t, \end{aligned}$$

(B1)

where $x_t$ is the smooth map defining the embedding and $Y_t$ is a vector field along $S_t$, a (not necessarily normal) section of TM restricted to $S_t$. As usual, if $\nu _t$ is the unit normal vector field along $S_t$, let $W=\nabla \nu _t$ be the shape operator of $S_t$, so the corresponding principal curvatures (the eigenvalues of W) are $\kappa _1$ and $\kappa _2$. Thus, the mean curvature is $H=\kappa _1+\kappa _2$ and the Gauss–Kronecker curvature is $K=\kappa _1\kappa _2$. For later reference, we recall that

$$\begin{aligned} {{\widetilde{K}}}=K-\frac{1}{2}\textrm{Ric}_g(\nu ,\nu ) \end{aligned}$$

is the modified Gauss–Kronecker curvature.

A well-known computation gives

$$\begin{aligned} \frac{d}{dt}A(t)\vert _{t=0}=\int _{ S} \textrm{div}_S YdS, \end{aligned}$$

(B2)

where A(t) is the area of $S_t$ and we agree to drop the subscript t upon evaluation at $t=0$. Next we decompose $Y_t$ into its normal and tangential components:

$$\begin{aligned} Y_t=f_t\nu _t+Y_t^\top , \quad f_t=\langle Y_t,\nu _t\rangle . \end{aligned}$$

(B3)

Thus, if we assume further that $S_t$ carries a boundary $\partial S_t$,

$$\begin{aligned} \frac{d}{dt}A(t)\vert _{t=0}= & {} \int _SfHdS+\int _S\textrm{div}_SY^\top dS\nonumber \\= & {} \int _SfHdS +\int _{\partial S}\langle Y,\mu \rangle d\partial S, \end{aligned}$$

(B4)

where $\mu$ is the outward unit normal vector field along $\partial S$ and we used that $H=\textrm{div}_S\nu$.

Let us assume now that M also carries a boundary, say $\Sigma$, with the variation being admissible in the sense that $\partial S_t\subset \Sigma$. It follows that $S=S_0$ is critical for the area under such variations satisfying the volume preserving condition

$$\begin{aligned} \int _SfdS=0 \end{aligned}$$

(B5)

if and only if the mean curvature is constant and S meets $\Sigma$ orthogonally along $\partial S$. We then say that S is a free boundary constant mean curvature (CMC) surface.

We now recall the corresponding notion of stability. Assuming that $S=S_0$ is a free boundary CMC as above, a well-known computation [59] gives the second variational formula for the area:

$$\begin{aligned} \frac{d^2 A}{dt^2}\vert _{t=0}=\int _{S}f\mathscr {L}_SfdS+\int _{\partial S} f\left( \frac{\partial f}{\partial \mu }-\kappa f\right) d\partial S, \end{aligned}$$

(B6)

where

$$\begin{aligned} {\mathscr {L}}_S=-\Delta _S -\left( \vert W\vert ^2+\textrm{Ric}_{g}(\nu ,\nu )\right) , \end{aligned}$$

(B7)

$\kappa =\langle \nu ,{\mathcal {W}} \nu \rangle$ and ${\mathcal {W}}=\nabla \eta$ is the shape operator of the embedding $\Sigma \hookrightarrow M$. Here, $\eta$ is the outward unit normal vector to M along $\Sigma$.

Recall that $S=S_0$ is strictly stable (as a free boundary CMC surface) if the right-hand side of (B6) is positive for any $f\ne 0$ satisfying (B5). Accordingly, we define

$$\begin{aligned} {\mathcal {F}}(S)=\left\{ f\in H^1(S);\int _SfdS=0\right\} . \end{aligned}$$

Proposition 23

A free boundary CMC surface S as above is strictly stable if and only if the first eigenvalue $\lambda _{{\mathscr {L}}_S}$ of the eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{ll} {\mathscr {L}}_Sf=\lambda f &{} \textrm{in}\,\,\, S, \\ \frac{\partial f}{\partial \mu }=\kappa f &{} \textrm{on}\,\,\, \partial S, \end{array} \right. \end{aligned}$$

is positive, where $f\in {\mathcal {F}} (S)$. Equivalently, for any $0\ne f\in {\mathcal {F}}(S)$,

$$\begin{aligned} \int _S\left( \vert \nabla _Sf\vert ^2-\left( \vert W\vert ^2+\textrm{Ric}(\nu ,\nu )\right) f^2\right) dS-\int _{\partial S}\kappa f^2d\partial S> 0. \end{aligned}$$

We now turn to the variational theory of the total mean curvature functional $\int _S H dS$. Here we assume that $S_t$ is closed ($\partial S=\emptyset$) and the variation is normal ($Y=f\nu$). A simple computation shows that the shape operator evolves as

$$\begin{aligned} \frac{\partial W}{\partial t}=-\nabla _S^2 f-(W^2+\textrm{Riem}_{g}^\nu )f, \end{aligned}$$

(B8)

where $\nabla _S^2$, the Hessian of f, is viewed as a (1, 1)-tensor, $\textrm{Riem}^\nu _g(\cdot )=\textrm{Riem}_g(\cdot ,\nu )\nu$ and ${{W^2=W\circ W}}$.

Proposition 24

In a Riemannian 3-manifold (M, g) as above, a closed surface extremizes the total mean curvature under volume (respectively, area) preserving variations if and only if ${{\widetilde{K}}}=\textrm{const}$ (respectively, ${{\widetilde{K}}}=\gamma H$, where $\gamma$ is a constant).

Proof

From $\partial dS_t/\partial t=fHdS_t$, the fact that the mean curvature evolves as

$$\begin{aligned} \frac{\partial H}{\partial t}={\mathscr {L}}_Sf, \end{aligned}$$

(B9)

and the algebraic identity $\vert W\vert ^2=H^2-2K$, we immediately see that

$$\begin{aligned} \frac{\partial }{\partial t}\int _{S_t}H dS_t\vert _{t=0}=2\int _S {{\widetilde{K}}} fdS_t, \end{aligned}$$

which proves the first statement. As for the second one, just combine the computation above with (B4) and take into account that $\partial S=\emptyset$. $\square$

In order to discuss the stability of this variational problem, we now compute the variation of ${{\widetilde{K}}}$. First, from $\partial \nu /\partial t=-\nabla _Sf$,

$$\begin{aligned} \frac{\partial }{\partial t}\textrm{Ric}_g(\nu ,\nu )=f(\nabla _\nu \textrm{Ric}_g)(\nu ,\nu )-2\textrm{Ric}_g({\nabla _Sf},\nu ). \end{aligned}$$

(B10)

As for the variation of K, we first recall the well-known formula

$$\begin{aligned} \frac{\partial }{\partial t}K=\textrm{tr}_S\left( \Pi \frac{\partial }{\partial t}W\right) , \end{aligned}$$

where $\Pi =HI-W$ is the Newton tensor [61]. Using (B8) we then get

$$\begin{aligned} \frac{\partial }{\partial t}K=-\textrm{tr}_S(\Pi \nabla _S^2f)-f\textrm{tr}_S(\Pi \, W^2)-f\textrm{tr}_S(\Pi \,\textrm{Riem}_g^\nu ). \end{aligned}$$

(B11)

To proceed we choose an orthonormal frame $e_A$, $A=1,2$, tangent to S with $(\nabla _S)_{e_A}e_B=0$ at the given point. We compute

$$\begin{aligned} \textrm{tr}_S(\Pi \nabla _S^2f)= & {} \Pi ^{AB}\langle (\nabla _S)_{e_A}\nabla _Sf,e_B\rangle \\= & {} \Pi ^{AB}e_A\langle \nabla _Sf,e_B\rangle -\Pi ^{AB}\langle \nabla _Sf,(\nabla _S)_{e_A}e_B\rangle \\= & {} (\Pi ^{AB}\nabla _Sf^B)_{;A}-\Pi ^{AB}_{\,\,\,\,;A}\nabla _Sf^B, \end{aligned}$$

where the semicolon denotes covariant derivation. By Codazzi equations, recalling that $h=g\vert _S$,

$$\begin{aligned} \Pi ^{AB}_{\,\,\,\,;A}= & {} (Hh^{AB})_{;A}-W^{AB}_{\,\,\,\,;A}\\= & {} (e_AH)h^{AB}-W^{AA}_{\,\,\,\,;B}-{{\langle R(e_A,e_B)\nu ,e_A\rangle }}\\= & {} e_BH-e_BH-\textrm{Ric}_{g}(\nu ,e_B)\\= & {} {{-\textrm{Ric}_{g}(\nu ,e_B)}}, \end{aligned}$$

so that

$$\begin{aligned} \textrm{tr}_S(\Pi \nabla _S^2f)=\textrm{div}_S(\Pi \nabla _Sf)+\textrm{Ric}_{g}(\nu ,\nabla _Sf). \end{aligned}$$

Thus, from (B11) and the algebraic identity $\textrm{tr}_S(\Pi W^2)=HK$, which holds in dimension 3,

$$\begin{aligned} \frac{\partial }{\partial t}K=-\Lambda _S f-\textrm{Ric}_{g}(\nu ,\nabla _Sf)-fHK-f\textrm{tr}_S(\Pi \textrm{Riem}_g^\nu ), \end{aligned}$$

where

$$\begin{aligned} \Lambda _S f=\textrm{div}_S(\Pi \nabla _Sf). \end{aligned}$$

(B12)

Together with (B10) this finally gives

$$\begin{aligned} \frac{\partial }{\partial t}{{\widetilde{K}}}=L_S f, \end{aligned}$$

(B13)

where

$$\begin{aligned} L_S =-\Lambda _S {{-HK-\frac{1}{2}(\nabla _\nu \textrm{Ric}_g)(\nu ,\nu )-\textrm{tr}_S(\Pi \textrm{Riem}_g^\nu )}} \end{aligned}$$

(B14)

is the corresponding Jacobi operator. We note that

$$\begin{aligned} -\int _Sf\Lambda _S{{\tilde{f}}}dS=\int _S\langle \Pi \nabla _S f,\nabla _S {{\tilde{f}}}\rangle dS, \end{aligned}$$

(B15)

for any functions f and ${{\tilde{f}}}$. In particular, $L_S$ is always self-adjoint. Moreover, it is easy to check that this operator is elliptic whenever $\Pi$ is positive definite.

We now consider a surface $S\subset M$ satisfying ${{\widetilde{K}}}=\mathrm{const.}$ and with the property that $\Pi$ is positive definite everywhere. We then say that S is strictly stable if

$$\begin{aligned} \frac{d^2}{dt^2}\int _{S}HdS\vert _{t=0}> 0, \end{aligned}$$

for any normal variation as in (B1) with $f\ne 0$. As before let us set

$$\begin{aligned} {\mathcal {G}}(S)=\left\{ f\in H^1(S);\int _S fdS=0\right\} . \end{aligned}$$

Proposition 25

S is strictly stable if and only if

$$\begin{aligned} \int _S\left( \langle \Pi \nabla _S f,\nabla _S f\rangle {{-f^2\left( HK+\frac{1}{2}(\nabla _\nu \textrm{Ric}_g)(\nu ,\nu )+\textrm{tr}_S(\Pi \textrm{Riem}_g^\nu )\right) }} \right) dS> 0, \end{aligned}$$

for any $0\ne f\in {\mathcal {G}}(S)$. Equivalently, the first eigenvalue $\lambda _{L_S}$ of the eigenvalue problem

$$\begin{aligned} L_Sf=\lambda f, \quad f\in {\mathcal {G}} (S), \end{aligned}$$

is positive.

Appendix C: The Gauss–Kronecker curvature in terms of the mean curvature

In this section we prove Proposition 13. Thus we aim to prove the identity (33) which expresses the Gauss–Kronecker curvature in terms of the mean curvature up to terms decaying fast enough at infinity. Our starting point is the fact that the radial vector field

$$\begin{aligned} X=(x_i-a_i)\frac{\partial }{\partial x_i} \end{aligned}$$

is conformal with respect to the Euclidean metric, i.e., $\mathcal L_X\delta =2\delta$, where ${\mathcal {L}}$ is the Lie derivative. From this we see that X is also conformal with respect to the metric $f_{m,c}^{\gamma _1, \gamma _2}\delta$ where

$$\begin{aligned} f_{m,c}^{\gamma _1, \gamma _2}=1+\frac{2m}{r}+\gamma _1\frac{c\cdot x}{r^3}+\gamma _2\frac{1}{r^2} \end{aligned}$$

for some $\gamma _1, \gamma _2\in {\mathbb {R}}$ and $c\in {\mathbb {R}}^3$. Indeed, there holds ${\mathcal {L}}_X(f_{m,c}^{\gamma _1, \gamma _2}\delta )=2\xi f_{m,c}^{\gamma _1, \gamma _2}\delta$, with

$$\begin{aligned} \xi (x)= & {} \frac{1}{f_{m,c}^{\gamma _1, \gamma _2}}\left( f_{m,c}^{\gamma _1, \gamma _2}+\frac{1}{2}\partial _kf_{m,c}^{\gamma _1, \gamma _2}(x_k-a_k)\right) \nonumber \\= & {} 1-\frac{m}{r}+\frac{2m^2-\gamma _2}{r^2}+{{\frac{m}{r^3} x\cdot a }}-{{\frac{\gamma _1}{r^3} x\cdot c }}+O(r^{-3})\nonumber \\= & {} 1-\frac{m}{\rho }+\frac{2m^2-\gamma _2}{\rho ^2}+{{\frac{2m}{\rho ^3} x\cdot a }}-{{\frac{\gamma _1}{\rho ^3} x\cdot c }}+O(\rho ^{-3}), \end{aligned}$$

(C1)

where in the last step we used that

$$\begin{aligned} r^k=\rho ^k+k\frac{(x-a)\cdot a}{\rho ^{2-k}}+O(\rho ^{-2+k}), \quad k\in {\mathbb {R}}. \end{aligned}$$

(C2)

Let us consider an aS metric of the form $g=f_{m,c}^{\gamma _1, \gamma _2}\delta +p$, where $p=O(r^{-2-\epsilon })$, which satisfies (18) with $\epsilon \ge 0$.

Proposition 26

The vector field X is almost conformal with respect to g in the sense that

$$\begin{aligned} {\mathcal {L}}_Xg=2\xi g+B, \quad \textrm{where}\quad B=O({\rho }^{-2-\epsilon }). \end{aligned}$$

(C3)

Proof

A direct computation shows that (C3) holds with $B={\mathcal {L}}_Xp-2\xi p$. Note however that

$$\begin{aligned} (\mathcal L_Xp)_{jk}=X^i\nabla _ip_{jk}+p_{ik}\nabla _jX^i+p_{ij}\nabla _kX^i, \end{aligned}$$

and the result follows given that $X=O(r)$. $\square$

We now take $\{e_1,e_2\}$ to be a local orthonormal frame on $S^2_\rho (a)$ and $\nu$ its outward unit normal vector. Recall that $W=\nabla \nu$ is the shape operator of $S^2_{\rho }(a)$ and $\Pi =HI-W$ denotes its Newton tensor. If $X^{\top }=X-\langle X,\nu \rangle \nu$ is the tangential component of X, then

$$\begin{aligned} ({\mathcal {L}}_{X^\top }g)(\Pi e_A,e_A)=(L_{X}g)(\Pi e_A,e_A)-2\langle X,\nu \rangle W(\Pi e_A,e_A), \end{aligned}$$

and we obtain from (C3) that

$$\begin{aligned} \langle \nabla _{\Pi e_A}X^\top ,e_A\rangle +\langle \nabla _{e_A}X^\top ,\Pi e_A\rangle =2\xi \langle \Pi e_A,e_A\rangle -2\langle X,\nu \rangle \langle W\Pi e_A,e_A\rangle +B(\Pi e_A,e_A). \end{aligned}$$

Since

$$\begin{aligned} \langle \nabla _{e_A}X^\top ,\Pi e_A\rangle =\langle e_A,\nabla _{\Pi e_A}X^\top \rangle , \end{aligned}$$

which is easily verified if we take the frame to be principal with respect to the shape operator W, this simplifies to

$$\begin{aligned} \langle \nabla _{e_A}X^\top ,\Pi e_A\rangle =\xi \langle \Pi e_A,e_A\rangle -\langle X,\nu \rangle \langle W\Pi e_A,e_A\rangle +\frac{1}{2}B(\Pi e_A,e_A). \end{aligned}$$

Thus, summing over A and using that $H_{a,\rho }^2-\vert W\vert ^2=2K_{a,\rho }$, we obtain

$$\begin{aligned} \sum _A \langle \nabla _{e_A}X^\top ,\Pi e_A\rangle =\xi H_{a,\rho }-2\langle X,\nu \rangle K_{a,\rho }+\frac{1}{2}\sum _A B(\Pi e_A,e_A). \end{aligned}$$

(C4)

In order to make use of this identity, which first appeared in [62], we need to determine the asymptotics of $X^\top$.

Proposition 27

One has $X^\top ={O(\rho ^{-1-\epsilon })}$.

Proof

Recalling that ${\mathfrak {r}}=(x-a)/\rho$, so that $X=\rho \mathfrak r_i\partial /\partial x_i$, one computes

$$\begin{aligned} \nu =\big (f_{m,c}^{\gamma _1, \gamma _2}\big )^{-1/2}\mathfrak r_i\frac{\partial }{\partial x_i}+O(\rho ^{-2-\epsilon }), \end{aligned}$$

(C5)

so that

$$\begin{aligned} \langle X,\nu \rangle =\big (f_{m,c}^{\gamma _1, \gamma _2}\big )^{1/2}\rho +O(\rho ^{-1-\epsilon }) \end{aligned}$$

(C6)

Thus,

$$\begin{aligned} \langle X,\nu \rangle \nu = \rho \,\mathfrak r_i\frac{\partial }{\partial x_i}+{O(\rho ^{-1-\epsilon })}, \end{aligned}$$

and the result follows. $\square$

We now observe that by (30) we may rewrite (54) as

$$\begin{aligned} \Pi =\frac{1}{2}H_{\rho ,a}I+O(\rho ^{-3}), \end{aligned}$$

so that

$$\begin{aligned} \frac{1}{2}\sum _AB(\Pi e_A,e_A)=\frac{1}{4}H_{\rho ,a}\textrm{tr}_{S^2_\rho (a)}B+O(\rho ^{-5}), \end{aligned}$$

where we used that ${{B=O(\rho ^{-2})}}$. Also, the left-hand side of (C4) may be treated similarly. Indeed, by Proposition 27,

$$\begin{aligned} \sum _A \langle \nabla _{e_A}X^\top ,\Pi e_A\rangle= & {} \frac{1}{2}H_{\rho ,a}\textrm{div}_{S^2_\rho (a)}X^\top +O(\rho ^{-5}). \end{aligned}$$

Putting all the pieces of our computation together and using (C6) we get

$$\begin{aligned} 2K_{a,\rho }= & {} \left( \frac{\xi }{\langle X,\nu \rangle }+\frac{1}{\langle X,\nu \rangle }\left( \frac{1}{4}\textrm{tr}_{S^2_\rho (a)}B-\frac{1}{2}\textrm{div}_{S^2_\rho (a)}X^\top \right) \right) H_{a,\rho }+O(\rho ^{-6})\\= & {} \left( \frac{\xi }{\langle X,\nu \rangle }+{ O(\rho ^{-3-\epsilon })}\right) H_{a,\rho }+O(\rho ^{-6}). \end{aligned}$$

The proof of Proposition 13 is completed if we note that by (C1) and (C6),

$$\begin{aligned} \frac{\rho \,\xi }{\langle X,\nu \rangle }= & {} \xi \left( \big (f_{m,c}^{\gamma _1, \gamma _2}\big )^{-1/2}+O(\rho ^{-2-\epsilon })\right) \\= & {} \left( 1-\frac{m}{\rho }+\frac{2m^2-\gamma _2}{\rho ^2}+\frac{2m}{\rho ^3}x\cdot a-\frac{\gamma _1}{\rho ^3}x\cdot c+ O(\rho ^{-3})\right) \\{} & {} \cdot \left( 1-\frac{m}{\rho }+\frac{3m^2-\gamma _2}{2\rho ^2}+\frac{m}{\rho ^3}x\cdot a-\frac{\gamma _1}{2\rho ^3}x\cdot c+ O(\rho ^{-2-\epsilon })\right) \\= & {} 1-\frac{2m}{\rho }+\frac{9m^2-3\gamma _2}{2\rho ^2}+\frac{3m}{\rho ^3}x\cdot a-\frac{3\gamma _1}{2\rho ^3}x\cdot c+ O(\rho ^{-2-\epsilon }). \end{aligned}$$

Appendix D: The proof of Proposition 19

Here we indicate how the argument in [26, Appendix F] may be used to prove Proposition 19. In fact, this method allows us to approach the problem in the category of manifolds considered in Definition 9.

Proposition 28

If (M, g) is an asymptotically flat 3-manifold with a non-compact boundary satisfying the RT condition then

$$\begin{aligned} \int _{S^2_{\rho ,+}(b)}\left( x_\alpha -b_\alpha \right) \left( H_{\rho ,+,b}-\frac{2}{\rho }\right) dS^{2,\delta ^+}_{\rho ,+}(b)=8\pi \mathfrak m \left( b_\alpha -C_\alpha ^+\right) +O(\rho ^{-\tau }), \quad \alpha =1,2.\qquad \end{aligned}$$

(D1)

Corollary 29

There holds

$$\begin{aligned} {\mathcal {C}}_\alpha ^+=-\lim _{\rho \rightarrow +\infty }\frac{1}{8\pi {\mathfrak {m}}} \int _{S^2_{\rho ,+}(\vec {0})}x_\alpha H_{\rho ,+,\vec {0}}dS^{2,\delta ^+}_{\rho ,+}(\vec {0}). \end{aligned}$$

The key ingredient in the proof is an integral identity derived from the fact that $S^2_{\rho ,+}(b)$ is a free boundary CMC surface with mean curvature $2/\rho$ with respect to the metric $\delta ^+$. In the following, for convenience we shall omit the area element of $S^2_{\rho ,+}(b)$ and the line element of $S^{1}_{\rho }(b):=\partial S^2_{\rho ,+}(b)$ in the respective integrals.

Proposition 30

There holds

$$\begin{aligned} \frac{1}{2}\int _{S^2_{\rho ,+}(b)}\left( x_\alpha -b_\alpha \right) e^+_{ij,k}{\mathfrak {r}}_i{\mathfrak {r}}_j{\mathfrak {r}}_k= & {} \int _{S^2_{\rho ,+}(b)}\left( x_\alpha -b_\alpha \right) \left( \frac{1}{2}e^+_{ij,k}{\mathfrak {r}}_j-2e^+_{ij}\frac{{\mathfrak {r}}_i{\mathfrak {r}}_j}{\rho }\right) \\{} & {} + \frac{1}{2}\int _{S^2_{\rho ,+}(b)}\left( e^+_{ii}{\mathfrak {r}}_\alpha +e^+_{i\alpha }{\mathfrak {r}}_i\right) \\{} & {} -\frac{1}{2}\int _{ S^{1}_{\rho }(b)}\left( x_\alpha -b_\alpha \right) e^+_{i3}{\mathfrak {r}}_i, \end{aligned}$$

where ${\mathfrak {r}}=(x-b)/\rho$.

Proof

Apply the identity that follows from equating the right-hand sides of (B2) and (B4) with $\mu =-\partial /\partial x_3$ to the vector field $Y_{(\alpha )}=(x_\alpha -b_\alpha ) e^+_{ij}\mathfrak r_i\partial /\partial x_j$ by taking into account that

$$\begin{aligned} \textrm{div}_{S^2_{\rho ,+}(b)}\,Y_{\alpha }= e^{+}_{i\alpha }\mathfrak r_i+\left( x_\alpha -b_\alpha \right) \left( \frac{e^+_{ii}}{\rho }-2\frac{e^+_{ij}}{\rho }\mathfrak r_i{\mathfrak {r}}_j+e^+_{ij,j}{\mathfrak {r}}_i-e^{+}_{ij,k}\mathfrak r_i{\mathfrak {r}}_j{\mathfrak {r}}_k\right) . \end{aligned}$$

$\square$

We now recall the expansion

$$\begin{aligned} H_{\rho ,+,b}-\frac{2}{\rho }=\frac{1}{2}e^+_{ij,k}\mathfrak r_i{\mathfrak {r}}_j{\mathfrak {r}}_k+2e^+_{ij}\frac{{\mathfrak {r}}_i\mathfrak r_j}{\rho }-e^{+}_{ij,i}{\mathfrak {r}}_j+\frac{1}{2}e^+_{ii,j}\mathfrak r_j-\frac{e^+_{ii}}{\rho }+E, \end{aligned}$$

where the remainder satisfies $E={O}(\rho ^{-1-2\tau })$ and $E^{({-1}')}=O(\rho ^{-2-2\tau })$; see [20, Lemma 2.1]. This reduces to (55) if we take $e^+=2mr^{-1}\delta ^++{O}(r^{-2})$, which provides the link between Propositions 28 and 19. It follows that

$$\begin{aligned} \int _{S^2_{\rho ,+}(b)}\left( x_\alpha -b_\alpha \right) \left( H_{\rho ,+,b}-\frac{2}{\rho }\right)= & {} -\frac{1}{2} \int _{S^2_{\rho ,+}(b)}\left( x_\alpha -b_\alpha \right) \left( e^+_{ij,i}-e^+_{ii,j}\right) {\mathfrak {r}}_j \\{} & {} +\frac{1}{2} \int _{S^2_{\rho ,+}(b)}\left( e^+_{i\alpha }{\mathfrak {r}}_i-e^+_{ii}{\mathfrak {r}}_\alpha \right) \\{} & {} -\frac{1}{2}\int _{ S^1_{\rho }(b)}\left( x_\alpha -b_\alpha \right) e^+_{i3}{\mathfrak {r}}_i + O({{\rho }}^{-\tau }), \end{aligned}$$

where Proposition 30 has been used to make sure that only those terms which are linear in ${\mathfrak {r}}$ survive in the right-hand side. We now observe that under the decay assumptions (including Regge–Teitelboim) the integrals

$$\begin{aligned} \int _{S^2_{\rho ,+}(b)}x_\alpha \left( e^+_{ij,i}-e^+_{ii,j}\right) \frac{b_{j}}{\rho }, \quad \int _{S^2_{\rho ,+}(b)} \left( e^+_{ij,i}-e^+_{ii,j}\right) \frac{b_{j}}{\rho }, \end{aligned}$$

and

$$\begin{aligned} \int _{S^2_{\rho ,+}(b)}\left( e^+_{i\alpha }\frac{b_i}{\rho }-e^+_{ii}\frac{b_\alpha }{\rho }\right) \end{aligned}$$

are $O(\rho ^{-\tau })$, the same happening to the boundary integrals

$$\begin{aligned} \frac{b_{\alpha }}{\rho }\int _{S^1_{\rho }(b)}x_\alpha e^+_{i3}, \quad \frac{b_{\alpha }b_i}{\rho }\int _{S^1_{\rho }(b)} e^+_{i3}. \end{aligned}$$

Thus, we end up with

$$\begin{aligned} \int _{S^2_{\rho ,+}(b)}\left( x_\alpha -b_\alpha \right) \left( H_{\rho ,+,b}-\frac{2}{\rho }\right)= & {} -\frac{1}{2} \int _{S^2_{\rho ,+}(b)}x_\alpha \left( e^+_{ij,i}-e^+_{ii,j}\right) \frac{x_{j}}{\rho }\\{} & {} +\frac{1}{2}\int _{S^2_{\rho ,+}(b)}\left( e^+_{i\alpha }\frac{x_i}{\rho }-e^+_{ii}\frac{x_\alpha }{\rho }\right) \\{} & {} -\frac{1}{2}\int _{S^1_{\rho }(b)}x_\alpha e^+_{i3}\frac{x_{i}}{\rho }\\{} & {} +\frac{1}{2} b_\alpha \int _{S^2_{\rho ,+}(b)} \left( e^+_{ij,i}-e^+_{ii,j}\right) \frac{x_{j}}{\rho }\\{} & {} +\frac{1}{2}b_\alpha \int _{S^1_{\rho }(b)} e^+_{i3}\frac{x_{i}}{\rho }+O(\rho ^{-\tau }). \end{aligned}$$

Comparing the right-hand side of the above with the definitions of ${\mathfrak {m}}$ and ${\mathcal {C}}^+$, the proof of Proposition 28, and hence of Proposition 19, follows.

Remark 11

The upshot of Corollary 29 is another expression for the center of mass ${\mathcal {C}}^+$, besides (28), derived from Hamiltonian methods, and the isoperimetric one appearing in Theorems 9 and 10. Another rendition of this invariant comes from [18, Theorem 2.4], this time in terms of certain asymptotic flux integrals involving the Einstein tensor of the metric in the interior and the Newton tensor along the boundary; see also [63]. It is remarkable indeed that this kind of invariant admits so many distinct manifestations.

Appendix E: The uniqueness of the free boundary CMC hemispheres

The very last piece of the argument leading to Theorem 10 uses the appropriate uniqueness of the free boundary CMC hemispheres in Theorem 9. Here we justify this step by following the reasoning in [5, Section 4]. We know from the analysis in Sect. 4 that for each $\rho$ large enough the corresponding hemisphere is a strictly stable free boundary CMC surface graphically described by a function $\phi _\rho$ on $S^2_{\rho ,+}({\mathcal {C}}^+)$ satisfying the bound

$$\begin{aligned} \Vert \rho ^{-1/2}\phi _\rho \Vert _{C^{2,\alpha }}^{(\rho )}\le C, \end{aligned}$$

where $C>0$ is an absolute constant and the weighted Hölder norm is defined as in the left-hand side of (51). The uniqueness claim is that, for $\rho$ large enough depending only on C, any other free boundary CMC hemisphere with the same mean curvature and which is graphed by a function satisfying this Hölder bound should coincide with (the graph of) $\phi _0:=\phi _\rho$. Indeed, assume there exists another such hemisphere, say associated to a function $\phi _1$. As in [5, Proposition 2.1], the asymptotic roundness of the graphs means that we may interpolate between the corresponding embeddings by setting

$$\begin{aligned} F_{t}(x)=F_{0}(x)+tu(x)\nu (x), \quad t\in [0,1], \end{aligned}$$

for some function $u(x)=\langle \vec {{\mathfrak {a}}},\nu (x)\rangle +q(x)$, where $\vec {{\mathfrak {a}}}\in {\mathbb {R}}^2$ is a vector and $q=O(\rho ^{-1})$. A crucial remark at this point is that all of these surfaces are free boundary (with a possibly non-constant mean curvature $H_{F_t}$ for $0<t<1$) and may be graphed by using functions satisfying the same Hölder bound as $\phi _0$. Since $H_{F_0}=H_{F_1}$, the variational vector field $Y=F_1-F_0$ satisfies

$$\begin{aligned} \vert Y\vert \le \Vert dH_{F_0}\Vert ^{-1}\sup _t\Vert d^2H_{F_t}(Y,Y)\Vert \le C_1\vert Y\vert ^2, \end{aligned}$$

where we used (70) applied to $dH_{F_0}=\mathscr {L}_{F_0}$, the Jacobi operator associated to $F_0$, and the fact that $\Vert d^2H_{F_t}\Vert =O(\rho ^{-3})$ uniformly in t. Thus, there exists an absolute constant $C_2>0$ such that $\vert Y\vert \le C_2$ implies $Y=0$. We next check that $\vert Y\vert$ (equivalently, $\vert \vec {{\mathfrak {a}}}\vert$) may be chosen small enough so as to fulfill this vanishing criterion if $\rho$ is large. We first note that, again because $H_{F_0}=H_{F_1}$,

$$\begin{aligned} \Vert dH_{F_0}Y\Vert \le \sup _{t}\Vert (dH_{F_t}-dH_{F_0})Y\Vert . \end{aligned}$$

(E1)

As in [64, Proposition 16] we compute that

$$\begin{aligned} dH_{F_t}Y={\mathscr {L}}_{F_t}u+Y^{\top }H_{F_t}, \end{aligned}$$

where $Y^\top$ is the tangential component of Y. Starting with (C5) we obtain $\vert Y^{\top }\vert =O(\rho ^{-3})$ and hence $\vert Y^{\top }H_{F_t}\vert =O(\rho ^{-4})$. Combining this with (64) we see that the right-hand side of (E1) is $O(\rho ^{-4})$. On the other hand, since $\langle \vec {{\mathfrak {a}}},\nu \rangle$ is an approximate eigenfunction of ${\mathscr {L}}_{F_0}$ under Neumann boundary condition with eigenvalue close to $6m\rho ^{-3}$, the left-hand side of (E1) is $\ge C_3\vert \vec {{\mathfrak {a}}}\vert \rho ^{-3}$. Thus, $\vert \vec {{\mathfrak {a}}}\vert \le C_4\rho ^{-1}$ and the uniqueness claim follows provided we take $\rho \ge C_2^{-1}C_4$.