Hence,

$$
\begin{align*}
\|v_k\|_{L^c(\mathbb{S}_1(0), \mathbb{R})} &= \left[ \int_{\mathbb{S}_1(0)} \left| \int_{SO(n)} \psi_k(g^{-1}) v(g \cdot x) d\mu(g) \right|^c d\mu^\sigma(x) \right]^{1/c} \\
&\le \left[ \int_{\mathbb{S}_1(0)} \int_{SO(n)} \psi_k(g^{-1}) |v(g \cdot x)|^c d\mu(g) d\mu^\sigma(x) \right]^{1/c} \\
&= \left[ \int_{SO(n)} \psi_k(g^{-1}) \int_{\mathbb{S}_1(0)} |v(g \cdot x)|^c d\mu^\sigma(x) d\mu(g) \right]^{1/c} \\
&= \left[ \int_{SO(n)} \psi_k(g^{-1}) \int_{\mathbb{S}_1(0)} |v(x)|^c d\mu^\sigma(x) d\mu(g) \right]^{1/c} \\
&= \|v\|_{L^c(\mathbb{S}_1(0), \mathbb{R})},
\end{align*}
$$

where we used the fact that the measure $d\mu^\sigma$ on $\mathbb{S}_1(0)$ is invariant under the action of $SO(n)$. Next, we claim that $v_k \in C^\infty(\mathbb{S}_1(0), \mathbb{R})$ and that $v_k \to v$ in $L^c(\mathbb{S}_1(0), \mathbb{R})$. For any $h \in SO(n)$, let $L_h$ denote the map from $\mathbb{S}_1(0)$ onto itself defined by $x \mapsto h \cdot x$. Then we have

$$
\begin{align*}
(L_h)^* v_k(x) &= v_k(h \cdot x) \\
&= \int_{SO(n)} \psi_k(g^{-1}) v(gh \cdot x) d\mu(g) \\
&= \int_{SO(n)} \psi_k(g^{-1}h^{-1}) v(g \cdot x) d\mu(g),
\end{align*}
$$

by the right invariance of the Haar measure. In particular, given $X \in \mathfrak{so}(n)$ and $t \in \mathbb{R}$, we can choose $h = \exp(tX)$. If $\Phi_t^X$ denotes the flow associated to the left invariant vector field $X$ on $G$, we know that $\Phi_{-t}^X(g) = g \exp(-tX) = gh^{-1}$ (see e.g. [33, Proof of Theorem 9.16]). Hence,

$$
\begin{align*}
(L_{\exp(tX)})^* v_k(x) &= \int_{SO(n)} \psi_k(\Phi_{-t}^X(g^{-1})) v(g \cdot x) d\mu(g) \\
&= \int_{SO(n)} ((\Phi_{-t}^X)^* \psi_k)(g^{-1}) v(g \cdot x) d\mu(g).
\end{align*}
$$

As a consequence,

$$
\begin{align*}
X v_k(x) &= \lim_{t \to 0} \frac{(L_{\exp(tX)})^* v_k(x) - v_k(x)}{t} \\
&= \int_{SO(n)} \left[ \lim_{t \to 0} \frac{((\Phi_{-t}^X)^* \psi_k)(g^{-1}) - \psi_k(g^{-1})}{t} \right] v(g \cdot x) d\mu(g) \\
&= \int_{SO(n)} (-X \cdot \psi_k)(g^{-1}) v(g \cdot x) d\mu(g)
\end{align*}
$$

is well-defined and continuous. Note that we can continue the argument adding
more and more vector fields on the left as the only property of $\psi_k$ used here is
that it is a smooth function. The convergence $v_k \to v$ in $L^c(\mathbb{S}_1(0), \mathbb{R})$ follows
now by standard arguments, using the fact that continuous functions are dense in
$L^c(\mathbb{S}_1(0), \mathbb{R})$ and are uniformly continuous as $\mathbb{S}_1(0)$ is compact (see e.g. [9, Theorem
4.22]).
We now claim that for any traceless symmetric tensor $T \in C^\infty(\mathbb{S}_1(0), \text{Sym}_2\mathbb{S}_1(0))$, we have

$$0 = \int_{\mathbb{S}_1(0)} v_k(x) \mathcal{D}^A \mathcal{D}^B T_{AB} d\mu^\sigma(x).$$

Indeed, from the fact that $SO(n)$ acts by isometries on $\mathbb{S}_1(0)$, for $u = \mathcal{D}^A \mathcal{D}^B T_{AB}$, we have

$$ \begin{aligned} \int_{\mathbb{S}_1(0)} u(x)v_k(x)d\mu^\sigma(x) &= \int_{\mathbb{S}_1(0)} u(x) \left( \int_{SO(n)} \psi_k(g^{-1})v(g \cdot x)d\mu(g) \right) d\mu^\sigma(x) \\ &= \int_{SO(n)} \psi_k(g^{-1}) \int_{\mathbb{S}_1(0)} u(x)v(g \cdot x)d\mu^\sigma(x)d\mu(g) \\ &= \int_{SO(n)} \psi_k(g^{-1}) \int_{\mathbb{S}_1(0)} u(g^{-1} \cdot x)v(x)d\mu^\sigma(x)d\mu(g) \\ &= \int_{SO(n)} \psi_k(g^{-1}) \int_{\mathbb{S}_1(0)} ((L_g)_*u)(x)v(x)d\mu^\sigma(x)d\mu(g). \end{aligned} $$

Again from the fact that $SO(n)$ acts by isometries on $\mathbb{S}_1(0)$, we have that

$$(L_g)_*u = (L_g)_*(\mathcal{D}^A \mathcal{D}^B T_{AB}) = \mathcal{D}^A \mathcal{D}^B ((L_g)_*T)_{AB}$$

is the double divergence of a symmetric trace free 1-tensor. Thus,

$$\int_{\mathbb{S}_1(0)} ((L_g)_*u)(x)v(x)d\mu^\sigma(x) = 0.$$

This shows that for any $T \in C^\infty(\mathbb{S}_1(0), \text{Sym}_2\mathbb{S}_1(0))$, we have

$$\int_{\mathbb{S}_1(0)} v_k(x) (\mathcal{D}^A \mathcal{D}^B T_{AB}) d\mu^\sigma(x) = 0.$$

Integrating by parts twice, this yields

$$\int_{\mathbb{S}_1(0)} \text{Hess}_{AB}^\sigma(v_k) T^{AB} d\mu^\sigma(x) = 0.$$

As $T$ is arbitrary, we conclude that $\text{Hess}^\sigma v_k \equiv 0$. From Proposition 5.6, we know that all $v_k$ live in the finite dimensional subspace $\mathcal{H} \subset L^c(\mathbb{S}_1(0), \mathbb{R})$. As $\mathcal{H}$ is closed, we can pass to the limit and conclude that $v \in \mathcal{H}$, showing that $v$ is smooth and satisfies (5.28) in the strong sense. $\square$

Note that Claim 5 implies that given $\mathcal{V} \in \mathcal{N}^g$, we can associate to it a unique function $v \in \mathcal{H}$. We denote this linear map by $\mathcal{B}: \mathcal{N}^g \to \mathcal{H}$. Recall (see a comment after Lemma 5.3) that $\mathcal{X}^\mu$ denotes the eignfunction of the Laplacian $\Delta^g$ that is asymptotic to the coordinate function $X^\mu$, $\mu = 0, 1, \dots, n$.

Claim 6. We have $\mathcal{B}(\mathcal{X}^0) = 1$ and, for all $i \in [1, n]$, $\mathcal{B}(\mathcal{X}^i) = x^i$. In particular, the map $\mathcal{B}$ is surjective.
Proof. We write $\mathcal{X}^i = X^i + (\mathcal{X}^i - X^i)$ and compute $h_u(r)$ with $\mathcal{V} = \mathcal{X}^i$. As $\mathcal{X}^i - X^i \in L_{\tau-1}^p(\mathbb{H}^n \setminus K, \mathbb{R})$ and $X^i = \sinh(r)x^i$ (see (5.11)), we conclude that

$$
\begin{align*}
\lim_{s \to \infty} h_u(s) &= \lim_{s \to \infty} \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} X^i(r, \theta) u(\theta) e^{-r} d\theta dr \\
&\quad + \lim_{s \to \infty} \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} (\mathcal{X}^i - X^i) u(\theta) e^{-r} d\theta dr \\
&= \lim_{s \to \infty} \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} x^i(\theta) \sinh(r) u(\theta) e^{-r} d\theta dr \\
&= \int_{\mathbb{S}_1(0)} x^i(\theta) u(\theta) d\theta \lim_{s \to \infty} \int_{s-1}^{s+1} \sinh(r) e^{-r} dr \\
&= \int_{\mathbb{S}_1(0)} x^i(\theta) u(\theta) d\theta,
\end{align*}
$$

observing that the limit in the second line is zero (see the proof of Claim 4 for a
similar argument). The proof for $\mathcal{X}^0$ is identical. $\square$

Before proving injectivity of $\mathcal{B}$ (which will be Claim 8), we need to obtain better estimates for the asymptotic behavior of functions in $\mathcal{N}^g$. We do this in the following two claims.

*Claim 7.* Let $f_2(s)$ be the function defined by

$$
f_2(s) := \int_{s-1}^{s+1} \int_{\mathbb{S}^{n-1}} (\partial_1(\mathcal{V}e^{-r}))^p d\theta \, dr.
$$

Then, $f_2(s) = O(s^p e^{-2ps})$.

*Proof.* We compute the derivative of $f_2$ with respect to $s$:

$$
\frac{d}{ds} f_2(s) = p \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} (\partial_1(\mathcal{V}e^{-r}))^{p-1} (\partial_1^2(\mathcal{V}e^{-r})) d\theta dr.
$$

Now note that

$$
\begin{align*}
\partial_1^2(\mathcal{V}e^{-r}) &= (\partial_1^2\mathcal{V})e^{-r} - 2(\partial_1\mathcal{V})e^{-r} + \mathcal{V}e^{-r} \\
&= e^{-r}(\text{Hess}_{11}\mathcal{V} + \Gamma_{11}^i\partial_i\mathcal{V} + \mathcal{V}) - 2(\partial_1\mathcal{V})e^{-r} \\
&= e^{-r}(\mathcal{V}g_{11} + \mathcal{T}_{11} + \Gamma_{11}^i\partial_i\mathcal{V} + \mathcal{V}) - 2(\partial_1\mathcal{V})e^{-r} \\
&= e^{-r}(2\mathcal{V} + \mathcal{T}_{11} + \Gamma_{11}^i\partial_i\mathcal{V} + \mathcal{V}e_{11}) - 2(\partial_1\mathcal{V})e^{-r} \\
&= -2\partial_1(\mathcal{V}e^{-r}) + e^{-r}(\mathcal{T}_{11} + \Gamma_{11}^i\partial_i\mathcal{V} + \mathcal{V}e_{11}).
\end{align*}
$$
Inserting this formula in that of the derivative of $f_2$, we get

$$ 
\begin{aligned}
\frac{d}{ds} f_2(s) &= -2p f_2(s) \\
&\quad + p \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} (\hat{\partial}_1(\mathcal{V}e^{-r}))^{p-1} e^{-r} (\mathcal{T}_{11} + \Gamma_{11}^i \hat{\partial}_i \mathcal{V} + \mathcal{V}e_{11}) d\theta dr \\
&\le -2p f_2(s) + p f_2(s)^{\frac{p-1}{p}} \left[ \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} |e^{-r} \mathcal{T}_{11}|^p d\theta dr \right)^{\frac{1}{p}} \right. \\
&\qquad \left. + \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} |\Gamma_{11}|_g^p |e^{-r} d\mathcal{V}|_g^p d\theta dr \right)^{\frac{1}{p}} \right. \\
&\qquad \left. + \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} e^{-pr} \mathcal{V}^p |e|_b^p d\theta dr \right)^{\frac{1}{p}} \right].
\end{aligned}
\quad (5.30) $$

From the proof of Claim 1 (see Equation (5.13)), we have

$$ \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} |e^{-r} \mathcal{T}_{11}|^p d\theta dr \right)^{\frac{1}{p}} \le \Lambda \|\mathcal{T}\|_{L_{\tau-1}^p} e^{-\lambda s}. $$

Also, from Claim 2, we have

$$ \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} e^{-pr} \mathcal{V}^p |e|_b^p d\theta dr \right)^{\frac{1}{p}} \lesssim \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} |e|_b^p d\theta dr \right)^{\frac{1}{p}} \\ \lesssim e^{-\lambda s}. $$

So the first and the third integrals in the right hand side of (5.30) are $O(e^{-\lambda s})$. Finally, for the second integral in (5.30), we remark that, from Claim 2, we have

$$ |e^{-r} d\mathcal{V}|_g \le |d(e^{-r} \mathcal{V})|_g + |\mathcal{V}| |de^{-r}|_g $$

and the right-hand side is bounded independently of $(r, \theta)$. So,

$$ \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} |\Gamma_{11}|_g^p |e^{-r} d\mathcal{V}|_g^p d\theta dr \right)^{\frac{1}{p}} \lesssim \left( \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} |\Gamma_{11}|_g^p d\theta dr \right)^{\frac{1}{p}} \\ \lesssim e^{-\lambda s}. $$

From (5.30), we conclude that, for some large constant $\Lambda$, we have

$$ \frac{d}{ds} f_2(s) \le -2p f_2(s) + \Lambda e^{-\lambda s} f_2(s)^{\frac{p-1}{p}}. $$

Equivalently, we have

$$ \frac{d}{ds} (f_2(s))^{1/p} \le -2 (f_2(s))^{1/p} + \frac{\Lambda}{p} e^{-\lambda s}. $$

As $\lambda \ge 2$, the estimate follows. $\square$

We now conclude the proof of Proposition 5.4 with the following claim:

*Claim 8.* The map $\mathcal{B}$ is injective.

Indeed, this claim combined with Claim 6 shows that $\dim(\mathcal{N}^g) = \dim(\mathcal{H}) = n+1$. Hence $\mathcal{N}^g = \operatorname{span}\{\mathcal{X}^0, \dots, \mathcal{X}^n\}$.
Proof of Claim 8. As $\mathcal{B}$ is linear, it suffices to study its kernel and show that it reduces to $\{0\}$. If $\mathcal{V} \in \ker \mathcal{B}$, we have that the function $v$ introduced in (5.27) is identically equal to zero.

From Claim 2, we have that the sequence of functions $\mathcal{U}_k(t, \theta) := \mathcal{V}(k+t, \theta)e^{-(k+t)}$ is uniformly bounded in $W^{1,p}(\mathcal{A}, \mathbb{R})$.

Passing to a subsequence $(\mathcal{U}_{\varphi(k)})_k$, where $\varphi$ is an increasing function, we can assume that $(\mathcal{U}_{\varphi(k)})_k$ converges strongly in $L^p(\mathcal{A}, \mathbb{R})$ and weakly in $W^{1,p}(\mathcal{A}, \mathbb{R})$ to some function $\mathcal{U}_{\infty}$. But note that, from Claim 7, we have $\|\partial_1 \mathcal{U}_{\varphi(k)}\|_{L^p(\mathcal{A})} \to 0$. So $\partial_1 \mathcal{U}_{\infty} = 0$ in the sense of distributions, meaning that $\mathcal{U}_{\infty}$ does not depend on $t$. Returning to the definition of $h_u(s)$ (Equation (5.23)), we have that, for any function $u \in L^c(\mathbb{S}_1(0), \mathbb{R})$,

$$0 = \lim_{k \to \infty} h_u(\varphi(k)) = \int_{[-1,1]} \int_{\mathbb{S}_1(0)} \mathcal{U}_{\infty}(t, \theta) u(\theta) d\theta dt.$$

So we conclude that $\mathcal{U}_{\infty} \equiv 0$, showing that

$$\lim_{k \to \infty} f_0(\varphi(k)) = \lim_{k \to \infty} \int_{-1}^{1} \int_{\mathbb{S}_1(0)} |\mathcal{U}_{\varphi(k)}(t, \theta)|^p d\theta dt = 0,$$

where the function $f_0$ was defined in Equation (5.10). We now show that $\mathcal{V}$ enjoys better decay than what we obtained in Claim 1. Note that the estimate we obtained in the proof of Claim 1 was fairly crude. Now we can improve it to the following:

$$\begin{aligned} \left| \frac{d}{ds} f_0(s) \right| &= \left| \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} \partial_1 [\mathcal{V}(r, \theta)^p e^{-pr}] d\theta dr \right| \\ &= p \left| \int_{s-1}^{s+1} \int_{\mathbb{S}_1(0)} (e^{-r}\mathcal{V})^{p-1} \partial_1 (\mathcal{V}e^{-r}) d\theta dr \right| \\ &\le p f_0(s)^{\frac{p-1}{p}} f_2(s)^{\frac{1}{p}}. \end{aligned}$$

This differential inequality can be rewritten as follows:

$$\left| \frac{d}{ds} f_0(s)^{1/p} \right| \le f_2(s)^{\frac{1}{p}} \lesssim s e^{-2s}.$$

From the fact that $f_0(\varphi(k)) \to 0$, we conclude that $f_0(s)$ tends to zero as $s$ goes to infinity and further that there exists a constant $\Lambda > 0$ such that

$$f_0(s) \le \Lambda s^p e^{-2ps}.$$

In particular,

$$\int_{r_0+1}^{\infty} e^{ps} f_0(s) ds < \infty. \qquad (5.31)$$

We rewrite this as follows:

$$\begin{aligned} \int_{r_0+1}^{\infty} e^{ps} f_0(s) ds &= \int_{s=r_0+1}^{\infty} e^{ps} \int_{r=s-1}^{s+1} \int_{\mathbb{S}_1(0)} |\mathcal{V}(r, \theta)|^p e^{-pr} d\theta dr ds \\ &= \int_{r=r_0}^{\infty} \int_{s=\max\{r_0+1, r-1\}}^{r+1} e^{ps} \int_{\mathbb{S}_1(0)} |\mathcal{V}(r, \theta)|^p e^{-pr} d\theta ds dr \\ &\ge \int_{r=r_0+2}^{\infty} \int_{s=r-1}^{r+1} e^{ps} \int_{\mathbb{S}_1(0)} |\mathcal{V}(r, \theta)|^p e^{-pr} d\theta ds dr \end{aligned}$$

where we have exchanged the order of integration with respect to the variables $s$ and $r$. Now the integral with respect to $s$ can be computed:

$$\int_{s=r-1}^{r+1} e^{ps} ds = \frac{e^p - e^{-p}}{p} e^{pr},$$
so

$$
\begin{align*}
\int_{r_0+1}^{\infty} e^{ps} f_0(s) ds &\geq \frac{e^p - e^{-p}}{p} \int_{r=r_0+2}^{\infty} \int_{S_1(0)} |\mathcal{V}(r, \theta)|^p d\theta dr \\
&\geq \frac{e^p - e^{-p}}{p} \int_{\mathbb{H}^n \setminus B_{r_0+2}(0)} |\mathcal{V}|^p \sinh(r)^{1-n} d\mu^b.
\end{align*}
$$

As a consequence, the estimate (5.31) implies that

$$
\int_{\mathbb{H}^n \setminus B_{r_0+2}(0)} |\rho^{\frac{n-1}{p}} \mathcal{V}|^p d\mu^b < \infty,
$$

i.e. that $\mathcal{V} \in L^{\frac{n-1}{p}}(M, \mathbb{R})$. In particular, if we choose $\tau' = -\frac{n-1}{p}$ and $q = p$ in Proposition 5.1, we have

$$
|\tau' + \frac{n-1}{p} - \frac{n-1}{2}| = \frac{n-1}{2} < \frac{n+1}{2}.
$$

So, the operator $P: u \mapsto -\Delta u + nu$ is an isomorphism from $W_{\tau'}^{2,p}(M, \mathbb{R})$ onto $L_{\tau'}^p(M, \mathbb{R})$. In particular, since, by elliptic regularity we have $\mathcal{V} \in W_{\tau'}^{2,p}(M, \mathbb{R})$, we deduce that $\mathcal{V} \in \ker(P) = \{0\}$. This ends the proof of Claim 8 and of Proposition 5.4. $\square$

Our next task is to show how Proposition 5.4 allows us to identify the hyperbolic
isometry in Theorem 5.2.

**Proposition 5.7.** There is a linear map $A_{\Phi_2 \circ \Phi_1^{-1}} : \mathcal{N}^b \to \mathcal{N}^b$ defined, for any $V \in \mathcal{N}^b$, by

$$
A_{\Phi_2 \circ \Phi_1^{-1}}(V) = V \circ \Phi_2 \circ \Phi_1^{-1} + \varepsilon(V),
$$

where $\varepsilon(V) \in W_{\tau^{-1}}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$. Further $A_{\Phi_2 \circ \Phi_1^{-1}}$ corresponds to a hyperbolic isometry, i.e. to an element $B \in O_\uparrow(n, 1)$ such that for all $V \in \mathcal{N}^b$ we have $A_{\Phi_2 \circ \Phi_1^{-1}}(V) = V \circ B + \varepsilon(V)$.

Proof. Lemma 5.3 defines a linear map $A_{\Phi_1} : \mathcal{N}^b \to \mathcal{N}^g$, where $V \in \mathcal{N}^b$ is mapped to the unique function $\mathcal{V} \in \mathcal{N}^g$ such that $\mathcal{V} = V \circ \Phi_1 + \text{l. o. t.}$, with correction terms belonging to $W_{\tau^{-1}}^{2,p}(M, \mathbb{R})$. Note that $A_{\Phi_1}$ is an injective map between two spaces of dimension $n+1$ so it is bijective. Similarly, we can define a map $A_{\Phi_2} : \mathcal{N}^b \to \mathcal{N}^g$. As a consequence, we get an invertible map $A_{\Phi_2 \circ \Phi_1^{-1}} : \mathcal{N}^b \to \mathcal{N}^b$ defined by

$$
A_{\Phi_2 \circ \Phi_1^{-1}} = A_{\Phi_1}^{-1} \circ A_{\Phi_2}.
$$

From the definition, we see that, given $V \in \mathcal{N}^b$, $A_{\Phi_2 \circ \Phi_1^{-1}}(V)$ is the unique function $V' \in \mathcal{N}^b$ such that $V' = V \circ \Phi_2 \circ \Phi_1^{-1} + \varepsilon(V)$ with $\varepsilon(V) \in W_{\tau^{-1}}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$, where $K'' := \mathbb{H}^n \setminus \Phi_1(M \setminus (K_1 \cup K_2))$.

We claim that $A := A_{\Phi_2 \circ \Phi_1^{-1}}$ is an isometry of $\mathcal{N}^b \simeq (\mathbb{R}^{n,1})^*$. Note that the lapse functions $V^\mu, \mu = 0, \dots, n$, satisfy

$$
(V^0)^2 - \sum_{i=1}^{n} (V^i)^2 = 1
$$

as they are nothing but the restrictions of the coordinate functions of $\mathbb{R}^{n,1}$ to the
hyperboloid, whereas for any $x \in \mathbb{H}^n$, we have

$$
1 = -\eta(x, x) = (V^0(x))^2 - \sum_{i=1}^{n} (V^i(x))^2.
$$
Composing with $\Phi_2 \circ \Phi_1^{-1}$, we get

$$ (V^0(\Phi_2 \circ \Phi_1^{-1}(x)))^2 - \sum_{i=1}^{n} (V^i(\Phi_2 \circ \Phi_1^{-1}(x)))^2 = 1. $$

Since for each $\mu$ we have $V^\mu \circ \Phi_2 \circ \Phi_1^{-1} = A(V^\mu) - \varepsilon(V^\mu)$ with $\varepsilon(V^\mu) \in W_{\tau-1}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$, we conclude that

$$ (A(V^0)(x))^2 - \sum_{i=1}^{n} (A(V^i)(x))^2 - 1 \in W_{\tau-2}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R}). $$

Now remark that $(A(V^0))^2 - \sum_{i=1}^n (A(V^i))^2$ is the restriction of a quadratic form over $\mathbb{R}^{n,1}$ to the hyperboloid. As the restriction of a polynomial function over $\mathbb{R}^{n,1}$ to $\mathbb{H}^n$ cannot belong to $W_{\tau-2}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$ unless the function is identically zero, we conclude that

$$ (A(V^0)(x))^2 - \sum_{i=1}^{n} (A(V^i)(x))^2 = 1 \qquad (5.32) $$

for any $x \in \mathbb{H}^n \setminus K'$. By analyticity, it follows that (5.32) holds for all $x \in \mathbb{H}^n$. Further, by homogeneity, we conclude that, identifying $\mathcal{N}^b$ with $(\mathbb{R}^{n,1})^*$ (see Proposition 2.3), we have

$$ (A(X^0))^2 - \sum_{i=1}^{n} (A(X^i))^2 = \eta(X, X), \qquad (5.33) $$

where $(X^0, X^1, \dots, X^n)$ denotes the standard basis of $(\mathbb{R}^{n,1})^*$. Let $B : \mathbb{R}^{n,1} \to \mathbb{R}^{n,1}$ be the transpose of $A$, i.e. the linear map such that

$$ \forall \mu = 0, \dots, n, \quad A(X^\mu) = X^\mu \circ B. $$

Equation (5.33) transforms into

$$ \eta(BX, BX) = (X^0 \circ B)^2 - \sum_{i=1}^{n} (X^i \circ B)^2 = \eta(X, X), $$

showing that $B$ is an isometry of $\mathbb{R}^{n,1}$ as claimed. The only thing left to be shown is that $B$ preserves time orientation. To prove this fact, note that $\eta$ induces a symmetric bilinear form $\eta^*$ on $\mathbb{R}^{n,1}$ defined by

$$ \eta^*(X^\mu, X^\nu) = \eta^{\mu\nu}. $$

The signature of $\eta^*$ is the same as that of $\eta$. In particular $\eta^*(X^0, X^0) < 0$ so $X^0$ belongs to one of the two connected components of time-like linear forms (i.e. forms $\varphi \in (\mathbb{R}^{n,1})^*$ such that $\eta(\varphi, \varphi) < 0$). As $X^0$ restricts to $V^0$ on $\mathbb{H}^n$ and all elements of this connected component can be reached by multiplying $X^0$ by a positive factor and composing it with an element of $O_\uparrow(n, 1)$, all elements in that connected component induce positive functions on $\mathbb{H}^n$, while elements in the other connected component of time-like linear forms induce negative functions on $\mathbb{H}^n$ and space-like linear forms induce linear forms changing sign over some totally geodesic hypersurface.

As $A(X^0)$ restricts to $V^0 \circ \Phi_2 \circ \Phi_1^{-1} + \text{h.o.t. over } \mathbb{H}^n \setminus K''$ and $V^0 \circ \Phi_2 \circ \Phi_1^{-1}$ is obviously positive on $\mathbb{H}^n \setminus K'', A(X^0)$ restricts to a positive function outside some compact set of $\mathbb{H}^n$, hence, by the previous classification of the lapse functions, $A(X^0)$ must belong to the same connected component of time-like forms as $X^0$. In particular,

$$ 0 < A(X^0) \left( \frac{\partial}{\partial X^0} \right) = X^0 \left( B \left( \frac{\partial}{\partial X^0} \right) \right), $$

showing that $B(\frac{\partial}{\partial X^0})$ remains a future-pointing vector (which must be time-like as $B$ is a Lorentzian isometry). $\square$
We now finish the proof of Theorem 5.2 with the following lemma:

**Lemma 5.8.** Assuming that $A_{\Phi_2 \circ \Phi_1^{-1}} = \text{Id}_{\mathbb{H}^n}$, we can write $\Phi_2 \circ \Phi_1^{-1}(x) = \exp_x \zeta(x)$ for some vector field $\zeta \in W_{\tau}^{2,p}(\mathbb{H}^n \setminus K'', T\mathbb{H}^n)$.

In the course of the proof of this lemma, we will need a result that we state here:

**Lemma 5.9.** For any symmetric 2-tensor $T$ defined on $M\setminus(K_1 \cup K_2)$, we have

$$ \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_1)^{-p\tau} [|\nabla T|_g^p + |T|_g^p] d\mu^g < \infty \\ \Leftrightarrow \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} [|D((\Phi_1)_*T)|_b^p + |((\Phi_1)_*T)|_b^p] d\mu^b < \infty. $$

*Proof.* Note that the metric $g$ (or more exactly $(\Phi_1)_*g$) is uniformly equivalent to the metric $b$, so

$$ \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_1)^{-p\tau} |T|_g^p d\mu^g \lesssim \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} |T|_b^p d\mu^b \\ \text{and} \\ \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} |T|_b^p d\mu^b \lesssim \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_1)^{-p\tau} |T|_g^p d\mu^g, $$

where we have chosen, here and in what follows, to write $T$ instead of $((\Phi_1)_*T)$ to avoid overloading the notation. Further, note that $\nabla T = DT + \Gamma \star T$, where $\Gamma \star T$ denotes a contraction of the Christoffel symbols as defined in (2.4) and the tensor $T$. Hence,

$$ \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_1)^{-p\tau} |\nabla T|_g^p d\mu^g \lesssim \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} |\nabla T|_b^p d\mu^b \\ \lesssim \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} [|DT|_b^p + |\Gamma \star T|_b^p] d\mu^b \\ \lesssim \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} [|DT|_b^p + |De|_b^p|T|_b^p] d\mu^b. $$

Assuming that the $W_{\tau}^{1,p}$-norm of $T$ defined with respect to the metric $b$ is finite:

$$ \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} [|DT|_b^p + |T|_b^p] d\mu^b < \infty, $$

we have, by the Sobolev injection, that $T \in L^{\infty}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$. Hence,

$$ \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_1)^{-p\tau} |\nabla T|_g^p d\mu^g \lesssim \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} |DT|_b^p d\mu^b \\ + \|T\|_{L^{\infty}(\mathbb{H}^n \setminus K'', S_2 M)} \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} |De|_b^p d\mu^b \\ < \infty. $$

Hence we have shown the following:

$$ \int_{\mathbb{H}^n \setminus K''} (\rho \circ \Phi_1)^{-p\tau} [|DT|_b^p + |T|_b^p] d\mu^b < \infty \\ \Rightarrow \int_{M\setminus(K_1 \cup K_2)} \rho^{-p\tau} [|\nabla T|_g^p + |T|_g^p] d\mu^g < \infty. $$

The converse implication is proven similarly. □
This lemma has the following important consequence. Applied to the chart $\Phi_2$ instead of $\Phi_1$, we have, due to the assumptions of Theorem 5.2,

$$ \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_2)^{-p\tau} [|\nabla(\Phi_2)_*e_2|_g^p + |(\Phi_2)_*e_2|_g^p] d\mu^g < \infty. $$

Note that we can safely replace $\rho \circ \Phi_2$ by $\rho \circ \Phi_1$ in this formula. Indeed, as explained in [22], both functions are uniformly comparable to $e^{-r}$, where $r$ is the distance function (with respect to the metric $g$) to any compact subset $K \neq \emptyset$ of $M$: in other words, there exists a constant $\Lambda > 0$ such that

$$ \Lambda^{-1} e^{-r(x)} \le (\rho \circ \Phi_1)(x) \le \Lambda e^{-r(x)} \text{ and } \Lambda^{-1} e^{-r(x)} \le (\rho \circ \Phi_2)(x) \le \Lambda e^{-r(x)}. $$

Hence,

$$ \int_{M\setminus(K_1 \cup K_2)} (\rho \circ \Phi_1)^{-p\tau} [|\nabla(\Phi_2)_*e_2|_g^p + |(\Phi_2)_*e_2|_g^p] d\mu^g < \infty. $$

Applying once again Lemma 5.9, we conclude that

$$ \int_{\mathbb{H}^n \setminus K''} \rho^{-p\tau} [|D(\Phi_2 \circ \Phi_1^{-1})_* e_2|_b^p + |(\Phi_2 \circ \Phi_1^{-1})_* e_2|_b^p] d\mu^b < \infty. $$

So, $(\Phi_2 \circ \Phi_1^{-1})_* e_2 \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$.

As a consequence, setting $\tilde{b} := (\Phi_2 \circ \Phi_1^{-1})_* b$, we have

$$ \begin{aligned} \tilde{b} - b &= (\Phi_2 \circ \Phi_1^{-1})_* b - b \\ &= (\Phi_2)_* ((\Phi_1)^* b - g) + (\Phi_2)_* g - b \\ &= (\Phi_2)_* (\Phi_1)^* (b - (\Phi_1)_* g) + e_2 \\ &= -(\Phi_2 \circ \Phi_1^{-1})_* e_1 + e_2. \end{aligned} $$

As both terms in the last line belong to $W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$, we have shown

$$ \tilde{b} - b \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n). \quad (5.34) $$

*Proof of Lemma 5.8.* Assuming that $A_{\Phi_2 \circ \Phi_1^{-1}} = \text{Id}_{\mathbb{H}^n}$, by Proposition 5.7 we can write

$$ V^{\mu} \circ \Phi_2 \circ \Phi_1^{-1} - V^{\mu} \in W_{\tau-1}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R}). $$

Now recall that the functions $V^{\mu}$, $\mu = 0, \dots, n$, provide a full set of coordinates for the standard embedding of the hyperbolic space into Minkowski spacetime. For any $x \in \mathbb{H}^n \setminus K'',$ we denote $X^{\mu}(x) = V^{\mu}(x)$ and $Y^{\mu} = V^{\mu}((\Phi_2 \circ \Phi_1^{-1})(x))$.

Applying Proposition 2.4 (see the comment after the Proposition) we find that $\Phi_2 \circ \Phi_1^{-1}(x) = \exp_x \zeta(x)$ where $\zeta(x)$ is given by

$$ \zeta(x) = f(u) (U - uX), $$

with $U = Y - X$, $u = \eta(U, U)/2$ and $f(u) = \frac{\ln(1 + u + \sqrt{2u + u^2})}{\sqrt{2u + u^2}}$. We have $U \in W_{\tau-1}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$. As a consequence, $u \in W_{2(\tau-1)}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$. As $p > n$ and $\tau + \frac{n-1}{p} \ge 2$, we have $\tau - 1 > 0$, showing that $u \in L^{\infty}(\mathbb{H}^n \setminus K'', \mathbb{R})$. As $f$ is analytic, we have $f(u) - 1 \in W_{2(\tau-1)}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$. All in all, we see that $\zeta \in W_{\mu_0}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{R}^{n,1})$ with $\mu_0 = \min\{\tau - 1, 2\tau - 3\}$. Since we assumed $\tau > 3/2$, we have $\mu_0 > 0$. Applying Proposition 2.6, we conclude that $\zeta \in W_{\mu_0}^{2,p}(\mathbb{H}^n \setminus K'', \mathbb{H}^n)$.

We will now show how to improve this rough estimate for $\zeta$. First, from Proposition 2.6, we get $\zeta \in W_{\mu_0}^{2,p}(\mathbb{H}^n \setminus K'', T\mathbb{H}^n)$. Further, from (5.34), we have $\tilde{b} - b \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$. We now use Equation (2.16). As we have $\mu_0 > 0$, we have

$$ c := c(|\zeta|^2) = \cosh(|\zeta|) = 1 + O(|\zeta|^2), \quad s := s(|\zeta|^2) = \frac{\sinh(|\zeta|)}{|\zeta|} = 1 + O(|\zeta|^2). $$
So $c-1, s-1 \in W_{2\mu_0}^{2,p}(\mathbb{H}^n\setminus K'', \mathbb{R})$. It then follows from the embedding

$$W_{\mu_0}^{1,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n) \hookrightarrow L_{\mu_0}^{\infty}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$$

and Equation (2.16) which we rewrite as

$$\mathcal{L}_{\zeta} b = \frac{1}{cs}(\tilde{b} - c^2b) - \frac{s}{c}(b(D.\zeta, D.\zeta) + \zeta^{\flat} \otimes \zeta^{\flat}) - \frac{1-s^2}{4cs|\zeta|^2}d|\zeta|^2 \otimes d|\zeta|^2 - \frac{1-cs}{2cs|\zeta|^2}[\zeta^{\flat} \otimes d|\zeta|^2 + d|\zeta|^2 \otimes \zeta^{\flat}],$$

that we have

$$\mathcal{L}_{\zeta} b = \tilde{b} - b + v,$$

where $v \in W_{2\mu_0}^{1,p}(\mathbb{H}^n\setminus K'', S_2\mathbb{H}^n)$. We now take the divergence with respect to $b$ of the previous equation:

$$\mathrm{div}(\mathcal{L}_{\zeta} b) = \mathrm{div}(\tilde{b} - b) + \mathrm{div}(v) \in L_{\min\{\tau, 2\mu_0\}}^{p}(\mathbb{H}^n\setminus K'').$$

Now note that the operator $\zeta \mapsto \mathrm{div}(\mathcal{L}_{\zeta} b)$ is elliptic and satisfies the assumptions of [32, Theorem C] with indicial radius $R = \frac{n+1}{2}$ $^6$. In particular, since $\zeta \in W_{\mu_0}^{2,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$, with $\mu_0 > 0$, we have

$$\mu_0 + \frac{n-1}{p} - \frac{n-1}{2} > -\frac{n+1}{2}.$$

As a consequence, [32, Proposition 6.5] implies that $\zeta \in W_{\mu_1}^{2,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$ with $\mu_1 = \min\{2\mu_0, \tau\} > \mu_0$. We can now repeat the argument using the improved estimate we have for $\zeta$ and get that $v \in W_{2\mu_1}^{1,p}(\mathbb{H}^n\setminus K'', S_2\mathbb{H}^n)$ showing that $\zeta \in W_{\mu_2}^{2,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$, with $\mu_2 = \min\{2\mu_1, \tau\}$. After a finite number of steps, we conclude that $\zeta \in W_{\tau}^{2,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$ as claimed. $\square$

*End of the proof of Theorem 5.2.* From Proposition 5.7, we know that there exists an element $B \in O_1(n, 1)$ such that, identifying $B$ with an isometry of the hyperbolic space, we have, for any $V \in \mathcal{N}_b$,

$$V \circ B = V \circ \Phi_2 \circ \Phi_1^{-1} + \varepsilon(V),$$

where $\varepsilon(V) \in W_{\tau-1}^{2,p}(\mathbb{H}^n\setminus K'', \mathbb{R})$. But $V \mapsto V \circ B^{-1}$ is an automorphism of $\mathcal{N}^b$, so we can replace $V$ by $V \circ B^{-1}$ in the previous estimate:

$$V = V \circ (B^{-1} \circ \Phi_2) \circ \Phi_1^{-1} + \varepsilon(V \circ B^{-1}),$$

where $\varepsilon(V \circ B^{-1}) \in W_{\tau-1}^{2,p}(\mathbb{H}^n\setminus K'')$. As a consequence, we can use Lemma 5.8 with $\Phi_2$ replaced by $B^{-1} \circ \Phi_2$:

$$(B^{-1} \circ \Phi_2) \circ \Phi_1^{-1}(x) = \exp_x \zeta(x)$$

for some vector field $\zeta \in W_{\tau}^{2,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$. Thus

$$\Phi_2 \circ \Phi_1^{-1}(x) = B(\exp_x \zeta(x))$$

with $\zeta \in W_{\tau}^{2,p}(\mathbb{H}^n\setminus K'', T\mathbb{H}^n)$. $\square$

<figure><img src="image_8.png" /></figure>

<footnote>$^6$The calculation of the indicial radius of the operator $\zeta \mapsto \mathrm{div}(\mathcal{L}_{\zeta} b)$, i.e. the vector Laplacian, can be found in [32, Proposition G] or [20, Lemme 6.1], it is easy to check that the replacement of $\mathcal{L}_{\zeta} b$ by $\mathcal{L}_{\zeta} b$ does not change the indicial roots.</footnote>
### 5.3. Covariance of the mass and the mass aspect function.
We are now equipped to prove the covariance of the mass aspect function. The strategy here is basically the same as the one outlined in [36], yet we are going to be more cautious about how we handle diffeomorphisms.

Assume given two charts at infinity $\Phi_1 : M\backslash K_1 \to \mathbb{H}^n\backslash K_1'$ and $\Phi_2 : M\backslash K_2 \to \mathbb{H}^n\backslash K_2'$ and set $e_1 := (\Phi_1)_*g - b$ (resp. $e_2 := (\Phi_2)_*g - b$). Assume further that $e_1, e_2 \in W_{\tau}^{1,p}(\mathbb{H}^n\backslash K'', S_2\mathbb{H}^n)$, where $p > n$ and $\tau > 3/2$ satisfies

$$ \tau + \frac{n-1}{p} \in \begin{cases} [2, n) & \text{if } n = 3, \\ (\frac{n}{2}, n) & \text{if } n > 3, \end{cases} $$

so that the assumptions of Theorem 5.2 are fulfilled (these assumptions will be made more restrictive in what follows). Then we can write $\Phi_{12} := \Phi_2 \circ \Phi_1^{-1} = A \circ \Phi_0$, where A is an isometry of the hyperbolic space and $\Phi_0(x) = \exp_x(\zeta(x))$ is a diffeomorphism asymptotic to the identity. We note that the lower bound

$$ \tau + \frac{n-1}{p} > \frac{n}{2} $$

is valid in dimension 3 implying, by [32, Lemma 3.6(b)], that $W_{\tau}^{1,p}(\mathbb{H}^n\backslash K'', S_2\mathbb{H}^n) \hookrightarrow W_{1/2}^{1,2}(\mathbb{H}^n\backslash K'', S_2\mathbb{H}^n)$ (see [32, Lemma 3.6(b)]), which guarantees that g is asymptotically hyperbolic of order 1/2 as in Definition 2.1.

To avoid repeating the formulas for the charge integrals, we denote the integrands in (4.5) and in (4.11) by

$$ \mathcal{U}(b, V, e, \bar{\chi}_k) := V (\mathrm{div}_b(e) - d \mathrm{tr}_b(e)) (-D^b \bar{\chi}_k) + \mathrm{tr}_b(e) dV(-D^b \bar{\chi}_k) - e(D^b V, -D^b \bar{\chi}_k), \quad (5.35) $$

where we highlight the reference metric b as it will subsequently be modified by diffeomorphisms, and where $V \in C_{-1}^{\infty}(\mathbb{H}^n, \mathbb{R})$ is an eigenfunction of the Laplacian as constructed in Proposition 3.6. So the mass integrals are given by

$$ p(e_1, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(b, V, e_1, \bar{\chi}_k) d\mu^b \quad \text{and} \quad p(e_2, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(b, V, e_2, \bar{\chi}_k) d\mu^b. $$

Since $g = (\Phi_1^{-1})_*(b + e_1)$, we have

$$ e_2 = (\Phi_{12})_*(b + e_1) - b = (\Phi_{12})_*(b + e_1 - \Phi_{12}^* b). $$

As a consequence, by invariance of the integral giving $p(e_2, V)$ under the diffeomorphism $\Phi_{12}$, we can write

$$ \begin{align*} p(e_2, V) &= \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(b, V, e_2, \bar{\chi}_k) d\mu^b \\ &= \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(\Phi_{12}^* b, \Phi_{12}^* V, \Phi_{12}^* e_2, \Phi_{12}^* \bar{\chi}_k) d\mu^{\Phi_{12}^* b} \\ &= \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(\Phi_{12}^* b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^* b, \bar{\chi}_k \circ \Phi_{12}) d\mu^{\Phi_{12}^* b}. \end{align*} $$

As the limit is independent of the choice of the cutoff functions $\bar{\chi}_k$ we choose, we can immediately replace $\bar{\chi}_k \circ \Phi_{12}$ by $\bar{\chi}_k$ in the last expression:

$$ p(e_2, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(\Phi_{12}^* b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^* b, \bar{\chi}_k) d\mu^{\Phi_{12}^* b}. \quad (5.36) $$

Now remark that, upon introducing the “intermediate” chart at infinity $\Phi_{1.5} := \Phi_0 \circ \Phi_1$ so that $\Phi_{1.5} \circ \Phi_1^{-1} = \Phi_0$ and $\Phi_2 \circ \Phi_{1.5}^{-1} = A$, we can split the discussion of the covariance of the mass aspect into two cases:
• The case $\Phi_{12} = A \in O_+(n, 1)$, for which $\Phi_{12}^*b = b$: In this case, Equation (5.36) immediately yields

$$p(e_2, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathbb{U}(b, V \circ \Phi_{12}, e_1, \bar{\chi}_k) d\mu^b = p(e_1, V \circ A).$$

• The case $\Phi_{12} = \Phi_0$ is a diffeomorphism asymptotic to the identity: This case is more complex. Our first task is to prove that we have

$$p(e_2, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathbb{U}(b, V, e_1 - \mathcal{L}_{\zeta} b, \bar{\chi}_k) d\mu^b,$$

removing most of the occurences of $\Phi_{12}^*b$. First note that $b - \Phi_{12}^*b \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n) \subset W_{1/2+\varepsilon}^{1,2}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n)$ for some constant $\varepsilon > 0$. Indeed, as we have

$$\tau + \frac{n-1}{p} > \frac{1}{2} + \frac{n-1}{2},$$

we can let $\varepsilon > 0$ be so small that the inequality

$$\tau + \frac{n-1}{p} > \frac{1}{2} + \varepsilon + \frac{n-1}{2}$$

remains true. Hence, from [32, Lemma 3.6(b)], we obtain that

$$W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n) \subset W_{1/2+\varepsilon}^{1,2}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n).$$

As we also have, from the Sobolev injection [32, Lemma 3.6(c)],

$$W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n) \subset L^{\infty}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n),$$

we see that the error terms can be estimated as follows:

$$\int_{\mathbb{H}^n} \left| \mathbb{U}(\Phi_{12}^*b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^*b, \bar{\chi}_k) \frac{d\mu^{\Phi_{12}^*b}}{d\mu^b} - \mathbb{U}(b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^*b, \bar{\chi}_k) \right| d\mu^b \\ \lesssim \|e_1 + b - \Phi_{12}^*b\|_{W_{1/2}^{1,2}} \|\Phi_{12}^*b - b\|_{W_{1/2}^{1,2}(\text{supp } D\bar{\chi}_k)} \\ \lesssim \|e_1 + b - \Phi_{12}^*b\|_{W_{1/2}^{1,2}} \|\Phi_{12}^*b - b\|_{W_{1/2+\varepsilon}^{1,2}} \max_{x \in \text{supp } D\bar{\chi}_k} \rho^{\varepsilon}(x) \\ \to_{k \to \infty} 0.$$

Next, it follows from Formula (2.16) that $b - \Phi_{12}^*b = -\mathcal{L}_{\zeta}b + O(|\zeta|^2 + |D\zeta|^2)$. So, by an argument similar to the previous one, we conclude that

$$\int_{\mathbb{H}^n} \left| \mathbb{U}(b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^*b, \bar{\chi}_k) - \mathbb{U}(b, V \circ \Phi_{12}, e_1 - \mathcal{L}_{\zeta}b, \bar{\chi}_k) \right| d\mu^b \\ \to_{k \to \infty} 0.$$

Replacing $V \circ \Phi_{12}$ by $V$ in (5.36) is allowed by the fact that $V \circ \Phi_{12} - V \in W_{\delta-1}^{1,p}(\mathbb{H}^n \setminus K'', \mathbb{R})$. The details of the argument are left to the reader. All in all, we have proven the following:

$$\int_{\mathbb{H}^n} \left| \mathbb{U}(\Phi_{12}^*b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^*b, \bar{\chi}_k) \frac{d\mu^{\Phi_{12}^*b}}{d\mu^b} - \mathbb{U}(b, V \circ \Phi_{12}, e_1 - \mathcal{L}_{\zeta}b, \bar{\chi}_k) \right| d\mu^b \\ \to_{k \to \infty} 0.$$

As a consequence, we have proven that, under the assumption that $e_1, e_2 \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n)$, and $\Phi_{12}$ is asymptotic to the identity, then if the limit

$$p(e_2, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathbb{U}(\Phi_{12}^*b, V \circ \Phi_{12}, e_1 + b - \Phi_{12}^*b, \bar{\chi}_k) d\mu^{\Phi_{12}^*b}$$
exists, we have

$$p(e_2, V) = \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{U}(b, V, e_1 - \mathcal{L}_{\zeta} b, \bar{\chi}_k) d\mu^b. \qquad (5.37)$$

Finally, we would like to get rid of the Lie derivative $\mathcal{L}_{\zeta} b$ in the above formula. This is essentially Bartnik's celebrated "curious cancellation" (see [7]), that is explained in [36]. Let $W = W_k$ denote the vector field defined by

$$(W_k)_i = S_{ij}(-D\bar{\chi}_k)^j, \quad \text{with } S_{ij} = (D_i\zeta_j - D_j\zeta_i)V + 2\zeta_i D_j V - 2\zeta_j D_i V.$$

Note that $S_{ij}$ is antisymmetric. Hence,

$$\begin{aligned}
\operatorname{div}(W) &= D^i W_i \\
&= (D^i S_{ij})(-D\bar{\chi}_k)^j - S_{ij} \operatorname{Hess}^{ij} \bar{\chi}_k \\
&= (D^i S_{ij})(-D\bar{\chi}_k)^j \\
&= [V D^i (D_i \zeta_j - D_j \zeta_i) + D^i V (D_i \zeta_j - D_j \zeta_i)] (-D\bar{\chi}_k)^j \\
&\quad + 2 [D^i \zeta_i D_j V + \zeta^i \operatorname{Hess}_{ij} V - D_i V D^i \zeta_j - \zeta_j \Delta V] (-D\bar{\chi}_k)^j \\
&= [V D^i (D_i \zeta_j + D_j \zeta_i) - 2V D^i D_j \zeta_i + D^i V (D_i \zeta_j - D_j \zeta_i)] (-D\bar{\chi}_k)^j \\
&\quad + 2 [D^i \zeta_i D_j V + \zeta^i T_{ij} - D_i V D^i \zeta_j] (-D\bar{\chi}_k)^j - 2(n-1)V\zeta_j,
\end{aligned}$$

where, as before, $\mathcal{T} := \operatorname{Hess} V - Vb$. Since

$$\begin{aligned}
2D^i D_j \zeta_i &= 2D_j D^i \zeta_i - 2R^l_{ij} \zeta_l \\
&= D_j(\operatorname{tr}(\mathcal{L}_{\zeta} b)) + 2\operatorname{Ric}^l_j \zeta_l \\
&= D_j(\operatorname{tr}(\mathcal{L}_{\zeta} b)) - 2(n-1)\zeta_j,
\end{aligned}$$

we obtain from Equation (5.35) that

$$\begin{aligned}
D^i W_i &= [V D^i (\mathcal{L}_{\zeta} b)_{ij} - V D_j (\operatorname{tr}(\mathcal{L}_{\zeta} b)) - D^i V (D_i \zeta_j + D_j \zeta_i)] (-D\bar{\chi}_k)^j \\
&\quad + [\operatorname{tr}(\mathcal{L}_{\zeta} b) D_j V + 2 \zeta^i T_{ij}] (-D\bar{\chi}_k)^j \\
&= \mathcal{U}(b, V, \mathcal{L}_{\zeta} b, \bar{\chi}_k) + 2 \zeta^i T_{ij} (-D\bar{\chi}_k)^j.
\end{aligned}$$

As $W$ has compact support, the integral of the divergence of $W$ vanishes. As a consequence, we conclude that Equation (5.37) becomes

$$\begin{aligned}
p(e_2, V) &= \lim_{k \to \infty} \int_{\mathbb{H}^n} [\mathcal{U}(b, V, e_1, \bar{\chi}_k) - \mathcal{U}(b, V, \mathcal{L}_{\zeta} b, \bar{\chi}_k)] d\mu^b \\
&= \lim_{k \to \infty} \int_{\mathbb{H}^n} [\mathcal{U}(b, V, e_1, \bar{\chi}_k) - \operatorname{div}(W) + 2\mathcal{T}(\zeta, -D\bar{\chi}_k)] d\mu^b \\
&= \lim_{k \to \infty} \int_{\mathbb{H}^n} [\mathcal{U}(b, V, e_1, \bar{\chi}_k) + 2\mathcal{T}(\zeta, -D\bar{\chi}_k)] d\mu^b. \qquad (5.38)
\end{aligned}$$

If $V \in \mathcal{N}$, we have $\mathcal{T} \equiv 0$ and we obtain

$$p(e_2, V) = p(e_1, V). \qquad (5.39)$$

In summary, we have proven:

**Theorem 5.10.** Assume given two charts at infinity $\Phi_1, \Phi_2$ for $(M, g)$ such that $e_1 := \Phi_{1*}g - b \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'_1)$ and $e_2 := \Phi_{2*}g - b \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K'_2)$, where $p > n$ and $\tau > 3/2$ satisfy

$$\tau + \frac{n-1}{p} \in \begin{cases} [2, n) & \text{if } n = 3, \\ (\frac{n}{2}, n) & \text{if } n > 3. \end{cases}$$
Set, as in the statement of Theorem 5.2, $\Phi_2 \circ \Phi_1^{-1} = A \circ \Phi_0$ where $A \in O_\uparrow(n, 1)$ is an isometry of $\mathbb{H}^n$ and $\Phi_0$ is a diffeomorphism asymptotic to the identity. Then we have, for any $V \in \mathcal{N}$

$$p(e_2, V) = p(e_1, V \circ A).$$

We will now focus on proving the covariance of the mass aspect function whenever the later is defined, see Proposition 4.2. To this end, we need to prove that the $\mathcal{T}$-term in (5.38) tends to zero as $k$ goes to infinity. We start by a refinement of Lemma 5.8.

**Lemma 5.11.** Under the assumptions of Theorem 5.10, assume further that $e_1, e_2 \in L_\delta^1(\mathbb{H}^n \setminus (K'_1 \cup K'_2), S_2\mathbb{H}^n)$ for some $\delta \in (-1, 1]$. If the charts $\Phi_1$ and $\Phi_2$ satisfy $\Phi_2 \circ \Phi_1^{-1}(x) = \exp_x \zeta(x)$ for some $\zeta \in W_\tau^{2,p}(\mathbb{H}^n \setminus K'')$, where $\tau$ is as in Theorem 5.10, we have $\zeta \in W_{\delta'}^{1,1}(\mathbb{H}^n \setminus K'', T\mathbb{H}^n)$ for any $\delta' < \delta$.

*Proof.* As $W_\tau^{1,p}(\mathbb{H}^n \setminus K'_1, S_2\mathbb{H}^n) \hookrightarrow L_\tau^\infty(\mathbb{H}^n \setminus K'_1, S_2\mathbb{H}^n)$, we have

$$e_1 \in L_\tau^\infty(\mathbb{H}^n \setminus K'_1, S_2\mathbb{H}^n) \cap L_\delta^1(\mathbb{H}^n \setminus K'_1, S_2\mathbb{H}^n).$$

By interpolation, we deduce that $e_1 \in L_\mu^q(\mathbb{H}^n \setminus K'_1, S_2\mathbb{H}^n)$ for any $q \in (1, \infty)$ and
$$\mu = \frac{1}{q}\delta + \left(1 - \frac{1}{q}\right)\tau.$$ Indeed, for $\lambda = 1/q$, we have

$$\begin{align*}
\|e_1\|_{L_{\mu}^q}^q &= \int_{\mathbb{H}^n \setminus K'_1} \rho^{-q\mu} |e_1|^q d\mu^b \\
&= \int_{\mathbb{H}^n \setminus K'_1} \rho^{-(q-1)\tau-\delta} |e_1|^{q-1} |e_1| d\mu^b \\
&\le \left( \sup_{\mathbb{H}^n \setminus K'_1} \rho^{-\tau} |e_1| \right)^{q-1} \int_{\mathbb{H}^n \setminus K'_1} \rho^{-\delta} |e_1| d\mu^b \\
&= \|e_1\|_{L_\tau^q}^{q-1} \|e_1\|_{L_\delta^1} \\
&< \infty.
\end{align*}$$

Similarly, $\|e_2\|_{L_\mu^q}(\mathbb{H}^n \setminus K'_2, S_2\mathbb{H}^n) < \infty$. Reasoning as in the proof of Lemma 5.8, we have

$$\mathcal{L}_\zeta b = \tilde{b} - b + v.$$

Here, however, note that the arguments that led to (5.34) can be straightforwardly modified to yield $\tilde{b} - b \in L_\mu^q(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$.

Further, due to the fact that $v$ is quadratic in $\zeta$ and $D\zeta$ with

$$\zeta, D\zeta \in L_{\tau}^{p}(\mathbb{H}^{n} \setminus K'', S_{2}\mathbb{H}^{n}) \cap L_{\tau}^{\infty}(\mathbb{H}^{n} \setminus K'', S_{2}\mathbb{H}^{n}),$$

we deduce that $v \in L_{2\tau}^{p/2}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$. As $\tau + \frac{n-1}{p} > \frac{n}{2}$, we have $2\tau + \frac{2(n-1)}{p} > n$. On the other hand, since $\delta + n - 1 \le n$, we conclude that

$$\mu + \frac{n-1}{q} = \tau \left(1 - \frac{1}{q}\right) + \frac{\delta + n - 1}{q} \le n.$$

Thus, we have $v \in L_\mu^q(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$ as long as $q \le p/2$. All in all, we conclude that, as long as $q \le p/2$, we have

$$\mathcal{L}_\zeta b \in L_\mu^q(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n).$$

Taking the divergence, we have $\mathrm{div}(\mathcal{L}_\zeta b) \in W_\mu^{-1,q}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$, where the negative exponent Sobolev spaces are defined in [1, Section 3.1]$^7$. From [1, Lemma B.2], we conclude that $\zeta \in W_\mu^{1,q}(\mathbb{H}^n \setminus K'', S_2\mathbb{H}^n)$.

<footnote>$^7$We chose not to give a precise definition of these function spaces, as they are only used in this proof.</footnote>
The last step is to convert this estimate for $\zeta$ into a $W^{1,1}$-estimate. We fix an arbitrary $\delta' < \delta$ and use again [32, Lemma 3.6]: the embedding

$$W_{\mu}^{1,q}(\mathbb{H}^n \setminus K_1' \cup K_2, S_2 \mathbb{H}^n) \hookrightarrow W_{\delta'}^{1,1}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n)$$

holds true provided that $q > 1$, which we already assumed, and

$$\mu + \frac{n-1}{q} > \delta' + \frac{n-1}{1}.$$

However, we have

$$\mu + \frac{n-1}{q} = \frac{1}{q}\delta + \left(1 - \frac{1}{q}\right)\tau + \frac{n-1}{q} = \frac{1}{q}\left(\delta + \frac{n-1}{1}\right) + \left(1 - \frac{1}{q}\right)\tau.$$

Hence, upon choosing $q$ close enough to 1, we can ensure that

$$\mu + \frac{n-1}{q} > \delta' + \frac{n-1}{1}.$$

This concludes the proof of the fact that $\zeta \in W_{\delta'}^{1,1}(\mathbb{H}^n \setminus K'', S_2 \mathbb{H}^n)$. $\square$

<figure><img src="image_9.png" /></figure>

We are now in a position to prove the covariance of the mass aspect function. It turns out that we cannot prove covariance for all functions $V \in \mathcal{E}^{\alpha}$, because of the slight loss of weight we had to impose in Lemma 4.5, but only for a restricted class $\mathcal{E}_0^{\alpha}$ of eigenfunctions $V$ that we now introduce.

The little Hölder space $c^{\alpha}(\mathbb{S}_1(0), \mathbb{R})$, $\alpha \ge 0$, is defined as the closure of $C^{\alpha}(\mathbb{S}_1(0), \mathbb{R})$ in $C^{\alpha}(\mathbb{S}_1(0), \mathbb{R})$. For non-integer $\alpha$, this space is strictly contained in $C^{\alpha}(\mathbb{S}_1(0), \mathbb{R})$. For example, if $\alpha \in (0, 1)$, $c^{\alpha}(\mathbb{S}_1(0), \mathbb{R})$ is defined as follows:

$$f \in c^{\alpha}(\mathbb{S}_1(0), \mathbb{R}) \iff \lim_{\epsilon \to 0} \sup_{\substack{x,y \in \mathbb{S}_1(0), \\ d(x,y) < \epsilon}} \frac{|f(x) - f(y)|}{d(x,y)^{\alpha}} = 0,$$

see e.g. [35, Section 1.1.2, Exercise 5]. A similar definition holds for larger values of $\alpha$. We define the set $\mathcal{E}_0^{\alpha}$ to be the set of eigenfunctions $V \in \mathcal{E}^{\alpha}$ such that $\rho V$ restricts to a function $v \in c^{\alpha}(\mathbb{S}_1(0), \mathbb{R})$ on the boundary of $\mathbb{H}^n$.

**Theorem 5.12.** Assume given two charts at infinity $\Phi_i$, $i = 1, 2$, for $(M, g)$ such that

$$e_i := \Phi_{i*}g - b \in W_{\tau}^{1,p}(\mathbb{H}^n \setminus K_i', S_2 \mathbb{H}^n) \cap L_{\delta}^{1}(\mathbb{H}^n \setminus K_i', S_2 \mathbb{H}^n),$$

where $p > n$ and $\tau > 3/2$ satisfy

$$\tau + \frac{n-1}{p} \in \begin{cases} [2, n) & \text{if } n = 3, \\ (\frac{n}{2}, n) & \text{if } n > 3, \end{cases}$$

and $\delta \in (-1, 1]$. Set $\Phi_2 \circ \Phi_1^{-1} = A \circ \Phi_0$ where $A \in O_\uparrow(n, 1)$ is an isometry of $\mathbb{H}^n$ and $\Phi_0$ is a diffeomorphism asymptotic to the identity. Let $V \in \mathcal{E}_0^{1-\delta}$ be an eigenfunction of the Laplacian, see Definition 3.7. Then we have

$$p(e_2, V) = p(e_1, V \circ A).$$

*Proof.* As mentioned before, the only missing step is to prove that, in the case of a more general eigenfunction $V$, the equality (5.39) pertains, i.e. that the $\mathcal{T}$-term in (5.38) gives no contribution in the limit $k \to \infty$.

We shall not prove (5.38) for all functions $V \in \mathcal{E}^{1-\delta}$ but instead only for those $V$ such that $v = \rho V$ is a $C^{1-\delta'}$ function on $\mathbb{S}_1(0)$ for some $\delta' \in [-1, 1)$, $\delta' < \delta$. Indeed, as $C^{1-\delta'}(\mathbb{S}_1(0), \mathbb{R})$ is dense in $c^{1-\delta}(\mathbb{S}_1(0), \mathbb{R})$ and the maps $P(e_1, \cdot)$ and $P(e_2, \cdot)$ are continuous linear forms over $C^{1-\delta}(\mathbb{S}_1(0), \mathbb{R})$ (see Proposition 4.3), if $P(e_1, \cdot)$ and $P(e_2, \cdot)$ agree on $C^{1-\delta'}(\mathbb{S}_1(0), \mathbb{R})$, they agree on the whole of $C^{1-\delta}(\mathbb{S}_1(0), \mathbb{R})$.
We claim that for a function $V$ as described, we have

$$ \lim_{k \to \infty} \int_{\mathbb{H}^n} \mathcal{T}(\zeta, -D\bar{\chi}_k) d\mu^b = 0. \qquad (5.40) $$

Indeed, it follows from Lemma 5.11 that $\zeta \in L_{\delta'}^1(\mathbb{H}^n \setminus K'', T\mathbb{H}^n)$. As $\mathcal{T} \in L_{\delta'}^{\infty}(\mathbb{H}^n, S_2\mathbb{H}^n)$, we have $|\mathcal{T}(\zeta, \cdot)| \in L^1(\mathbb{H}^n \setminus K'', \mathbb{R})$. Since the sequence of 1-forms $-D\bar{\chi}_k$ is bounded in $L^{\infty}(\mathbb{H}^n, T^*\mathbb{H}^n)$ and converges a.e. to zero, we conclude, by the dominated convergence theorem, that (5.40) holds. $\square$

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(R. Gicquaud) INSTITUT DENIS POISSON, UNIVERSITÉ DE TOURS, PARC DE GRANDMONT, 37200 TOURS, FRANCE

Email address: romain.gicquaud@idpoisson.fr

(A. Sakovich) DEPARTMENT OF MATHEMATICS, UPPSALA UNIVERSITY, SWEDEN

Email address: anna.sakovich@math.uu.se