Therefore it suffices to prove $(\Phi, \mathcal{N})=0$. Recall that $\mathcal{N}$ satisfies the wave equation (<ref>) and since $\mathcal{N}$ is also a mode, we can rewrite it as

\begin{equation}
(\weight\pa_r)^2\mathcal{N}+(\sigma^2-\frac{\weight k^2}{r^2})\mathcal{N}=-\frac{2\BQ J}{r^2}=-\frac{\weight\BQ(X+Y)}{2r^2}.
\end{equation}

Using (<ref>) and (<ref>), we find

\begin{align*}
X+Y&=-(\frac{P_{X_0}}{2rH^2}+\frac{P_{Y_0}}{2rH^2	})\Phi+(\frac{P_{X_1}}{2H}+\frac{P_{Y_1}}{2	H})\Phi'-(\frac{2\BQ P_{X\mathcal{N}}}{r^2H^2}+\frac{2\BQ P_{Y\mathcal{N}}}{r^2H^2})\mathcal{N}\\
&=-(\frac{H}{r}-\frac{P_Z}{rH})\Phi-2\weight\Phi'-\frac{16\BQ \weight}{r^2H}\mathcal{N}
\end{align*}

and thus (see <cit.> and <cit.>)

\begin{equation}\label{eq:eqforN2}
\end{equation}

\begin{equation}\label{eq:VFN}
V_{\mathcal{N}}=\frac{\weight k^2}{r^2}+\frac{8\BQ^2\weight^2}{r^4 H},\quad F_{\mathcal{N}0}=\frac{\BQ \weight}{2r^2}(\frac{H}{r}-\frac{P_Z}{rH}),\quad F_{\mathcal{N}1}=\frac{\BQ \weight^2}{r^2}.
\end{equation}

Now we have two coupled second order ODEs (<ref>) and (<ref>)for $\Phi$ and $\mathcal{N}$ and in fact we can transform them into two decoupled second order ODEs by introducing suitable linear combinations of $\Phi$ and $\mathcal{N}$ (see <cit.> and <cit.>). Concretely, we set

\begin{equation}\label{eq:defofPsi+-}
\Psi_{\pm}=a_{\pm}\Phi+b_{\pm}\mathcal{N}
\end{equation}

where $a_{\pm}, b_{\pm}$ are smooth on $\hX$ (which extends a little bit beyond the event horizon $r=\ehRN$)

\begin{equation}\label{eq:a+-b+-}
(a_+, b_+)=(\frac{\BQ m}{2\tilde{c}}+\frac{\BQ}{2r}, 1),\quad (a_-, b_-)=(\frac{\tilde{c}}{6\Bm}-\frac{2\BQ^2}{3\Bm r}, -\frac{4\BQ}{3\Bm}),\quad \tilde{c}=3\Bm+\sqrt{9\Bm^2+4\BQ^2 m}.
\end{equation}

Since $\begin{vmatrix}
\frac{\BQ m}{2\tilde{c}}+\frac{\BQ}{2r}&1\\
\frac{\tilde{c}}{6\Bm}-\frac{2\BQ^2}{3\Bm r}& -\frac{4\BQ}{3\Bm}
\end{vmatrix}=-\frac{2\BQ^2m}{3\Bm \tilde{c}}-\frac{\tilde{c}}{6\Bm}<0$, we can recover $\Phi, \mathcal{N}$ from $\Psi_{\pm}$. Next the decoupled equations for $\Phi_{\pm}$ are given as

\begin{equation}\label{eq:eqforPsi+-}
\end{equation}

\begin{equation}\label{eq:defofV+-}
\end{equation}

Introducing the tortoise coordinate $r_*=\int \weight^{-1}\,dr$, we see that the equations for $\Psi_{\pm}$ are reduced to an eigenvalue problem of the type $A\Psi=\sigma^2\Psi$ on the region $r^*\in(-\infty,\infty)$ where $A=-\pa_{r^*}^2+V(r)$ is a self-adjoint operator. As explained in <cit.>, we expect $V(r)$ to be non-negative on $r\in[\ehRN, \infty)$ and then naturally so is $A$.

Given $r\in [\ehRN, \infty)$, we have $V_{+}\geq0$ for $l\geq 1$, i.e., $k^2\geq 2$ and $m\geq0$.

Since $\ehRN=\Bm+\sqrt{\Bm^2-\BQ^2}$ and $H=m+\frac{2}{r}(3\Bm-\frac{2\BQ^2}{r})$, in the subextremal case $\BQ<\Bm$, we have $r\geq \ehRN>\Bm>\BQ>\frac{2\BQ^2}{3\Bm}$ and thus $H(r)>0$. We first analyze

\[
\]

since $x=\frac{2\Bm}{r}\in[0, 2), z\geq 0, m\geq0$, we have $-3x^2+(2z+6)x+m(m+4)\geq m(m+4)\geq 0$ and thus $F_{\Phi}\geq0$. Next it is clear from (<ref>) that $V_{\mathcal{N}}\geq 0$. Therefore $V_+\geq0$.

However, the positivity of $V_-$ fails. To deal with this issue, we follow <cit.> and introduce the procedure S-deformation of $V_-$. More specifically, suppose $\Psi$ is compactly supported in $r_*$, we have
\[(\Psi, A\Psi)_{L^2(dr_*)}=\int \abs{\pa_{r_*}\Psi}^2+V(r)\abs{\Psi}^2\,dr_*=\int \abs{\wt{D}\Psi}^2+\wt{V}(r)\abs{\Psi}^2\,dr_*\]

where $\wt{D}=\pa_{r_*}+S=\weight\pa_r+S$ and $\wt{V}=V+\weight\pa_{r}S-S^2$. We apply the $S$-deformation to $V_-$ with

\[
S=S_-:=\frac{\weight\pa_r H_-}{H_-}=\weight\pa_r(\log H_-)\!\quad\!\mbox{where}\!\quad\! H_-=k^2-2+\frac{\tilde{c}}{r}>0 \!\quad\!\mbox{on}\!\quad\![\ehRN, \infty)\!\quad\!\mbox{for}\!\quad\! k^2\geq 2(l\geq1)
\]

and then

\begin{equation}\label{eq:Sdedor}
\wt{V}_-=\frac{k^2(k^2-2)\weight}{r^2H_-}\geq 0\!\quad \!\mbox{on}\!\quad \![\ehRN, \infty)\quad\mbox{for}\quad k^2\geq 2(l\geq 1).
\end{equation}

Namely, with $H_-$ and $V_-$ (which are also smooth near $r=\ehRN$) defined above, we write $\wt{D}=\pa_{r_*}+S_-=e^{-\int\! S_-dr_*}\pa_{r_*}e^{\int\!S_-dr^*}=H_-^{-1}\pa_{r_*}H_-=H_-\weight\pa_rH_-$ and find that $D^*$ which is the adjoint of $D$ with respect to $dr_*=\weight^{-1}dr$ can be expressed as $-H_-\pa_{r_*}H_-^{-1}=-H_-\weight\pa_rH_-^{-1}$. As a consequence, we have

\begin{equation}\label{eq:Psi-}
\begin{split}
\wt{D}^*\wt{D}\Psi_-+\wt{V}_-\Psi_-&=-H_-\weight\pa_r\Bigl(H_-^{-2}\weight\pa_r(H_-\Psi)\Bigr)+\wt{V}_-\Psi\\
&=-(\weight\pa_r)^2\Psi-\Psi H_-\weight\pa_r(H_-^{-2}\weight\pa_rH_-)+(V+H_-^{-1}(\weight\pa_r)^2H_--2H_-^{-2}(\weight\pa_rH_-)^2)\Psi\\
\end{split}
\end{equation}

Now we are at the position to show that $\Psi_{\pm}=0$ on $r>\ehRN$, and thus from the previous discussion the mode solution $(\dg, \dA)$ with $\sigma\neq0$ is a pure gauge solution. To achieve this, we first need to discuss the behavior of $\Psi_{\pm}$ near $r=\ehRN$ and $r=\infty$, i.e., $r_*=\mp\infty$. Since $H, a_{\pm}, b_{\pm}$ are bounded away from zero, we only need to analyze the contributions of $X, Y,Z,\mathcal{N}$. Note that $\wt{F}, J$ are smooth on $\hX$ (which extends a little bit beyond the event horizon). Therefore, the contribution of $J=(X+Y)/4$ to $\Phi$ is smooth on $\hX$. As for the contribution from $Z$, we notice that $\pa_t, \pa_{r_*}$, which are expressed as $\pa_t, \weight\pa_r$ in the static coordinates $(t,r)$, are smooth vector fields on $\hX$, and thus $Z=\weight\wt{F}_{tr}=-\wt{F}(\pa_t, \pa_{r^*})$ is smooth on $\hX$ as well. As a result, $\Phi=e^{-i\sigma t_*}C^{\infty}([\ehRN, \infty)_r)$, which is written in the static coordinates as

\[
\Phi=(t, r)=e^{-i\sigma t}\Theta(r)\quad\mbox{where}\quad \Theta(r)\in e^{\pm i\sigma r_*}C^\infty([\ehRN, \infty)) \quad \mbox{when}\quad r^*\to\pm\infty.
\]

which implies the exponential decay of $\Phi$ as $r_*\to-\infty$ when $\IM\sigma>0$. Also, when $r_*\to\infty$, since $\dg=e^{-i\sigma t_*}\dg_0$ with $\dg_0\in\eHb^{\infty, \ell}(\CX;\scsym)$, we still obtain the rapid decay of $\Phi$ when $r^*\to\infty$. As for $\mathcal{N}$, from the previous discussion, we know $\mathcal{N}$ is a mode, that is, $\mathcal{N}=e^{-i\sigma t_*}\mathcal{N}_0$ with $\mathcal{N}_0\in C^\infty([\ehRN, \infty)_r)\cap \eHb^{\infty, \ell}(\CX)$ as well, and thus decays rapidly when $r_*\to\pm\infty$. Consequently, $\Psi_{\pm}$ decays rapidly as well when $r_*\to\pm\infty$ when $\IM \sigma>0$.

Then when $\IM \sigma>0$, by pairing $(-\pa_{r_*}^2+V_+-\sigma^2)\Psi_+=0$ and $(-\wt{D}^*\wt{D}+\wt{V}_--\sigma^2)\Psi_-=0$ with $\overline{\Psi}_+$ and $\overline{\Psi}_-$ respectively with respect to the $L^2(\BR_{r_*}; dr_*)$ and then integrating by parts (whose validity is guaranteed by the rapid decay of $\Psi_{\pm}$ when $r_*\to\pm\infty$ ), one obtains

\[
0=\norm{\pa_{r_*}\Psi_+}_{L^2}+\norm{V_+^{1/2}\Psi_+}_{L^2}-\sigma^2\norm{\Psi_+}_{L^2},\quad 0=\norm{\wt{D}\Psi_-}_{L^2}+\norm{\wt{V}_-^{1/2}\Psi_-}_{L^2}-\sigma^2\norm{\Psi_-}_{L^2}.
\]

When $\RE\sigma=0$ and thus $\sigma^2<0$, we have a sum of squares and thus $\Psi_{\pm}=0$. When $\RE \sigma\neq0$, the imaginary part $-2i\RE\sigma\IM\sigma\norm{\Psi_{\pm}}_{L^2}=0$ and thus $\Psi_{\pm}=0$ on $(\ehRN, \infty)$.

When $\sigma\in \BR\setminus\{0\}$, we still have $e^{i\sigma t}\Phi, e^{i\sigma t}\mathcal{N}\in e^{\pm i\sigma r_*}\Bigl(C^\infty([\ehRN, \infty))\cap\eHb^{\infty, \ell}(\CX)\Bigr)$ when
$r_*\to\pm\infty$ for some $\ell\in\BR$, and thus so do $\Psi_{\pm}$. First, by normal operator arguments as in the proof of Proposition <ref>, we find $\Psi_+=e^{-i\sigma t}e^{i\sigma r_*}(a+\mathcal{A}^{1-})$ with $a\in\BC$ when $r_*\to\infty$. Next, a standard boundary pairing argument, see the proof of Theorem <ref> below for the detail (starting at (<ref>)), allows us to conclude $\lim _{r_*\to\pm\infty}e^{\mp i\sigma r_*}\Psi_{+}=0$. Then an indicial root argument implies $\Psi_+\in e^{-i\sigma t}e^{i\sigma r_*}\mathcal{A}^\infty$, i.e., $\Psi_+$ vanishes to infinite order at $r_*=\infty$. Lastly, the unique continuation at infinity (<cit.>) implies $\Psi_+=0$ on $(\ehRN, \infty)$. The proof for $\Psi_-$ proceeds similarly.

Next we treat the stationary mode solutions, i.e., $\sigma=0$. We still assume $l\geq2$. Now $(\dg, \dA)$ and thus $\wt{F}, J, N$ and $X, Y,Z$ are stationary.

We first analyze the nature of $\mathcal{N}$ defined as $\hstar N=\hd\mathcal{N}$. Let $\hstar N=N_0dt_*+N_1dr$ where $N_0, N_1$ are smooth functions of $r$ only, and then $\hd\hstar N=0$ gives $\pa_rN_0=0$ and thus $N_0$ is a constant. Therefore, we have the unique (up to additive constants) $\mathcal{N}=N_0t_*+\int_{r_-}^r N_1(s)ds$, which is a generalized mode with frequency $\sigma=0$ and grows at most linearly in $t_*$. Since $X,Y,Z$ are stationary, equation (<ref>) implies

\begin{equation}\label{eq:Z=0}
\end{equation}

which in turn implies $\pa_t\mathcal{N}=0$ by using equation (<ref>). Therefore, we conclude $N_0=0$ and $\mathcal{N}$ is stationary as well.

Alternatively, using $N\in\eHb^{\ell, \infty}(\CX; \scform)\subset \mathcal{A}^{\ell+3/2}\subset\mathcal{A}^{0+}$ (because $-3/2<\ell<-1/2$) when $\sigma=0$, it also follows that $N_0=c=0$.

As for $\Phi$, we notice that $Z/(i\sigma)$ is not well-defined when $\sigma=0$ and we need to do some remedies. Using (<ref>), $Z/(i\sigma)$ can be expressed as a linear combination of $X, Y,\mathcal{N}, \mathcal{N}'$ and thus we can rewrite $\Psi_{\pm}$ as linear combinations of $X,Y,\mathcal{N},\mathcal{N}'$ whose coefficients depend on $\sigma$

\begin{equation}\label{eq:newexpressionforPsi+-l2}
\Psi_{\pm}(\sigma)=C_{X\pm}(\sigma)X+C_{Y\pm}(\sigma)Y+C_{\mathcal{N}\pm}(\sigma)\mathcal{N}+C_{\mathcal{N}'\pm}(\sigma)\mathcal{N}'
\end{equation}

where $C_{\bullet \pm}(
\sigma)$ are rational functions of $r$ depending on $\sigma\in\BC$. To make the dependence on $\sigma$ explicit, we write (see <cit.>)

\begin{equation}\label{eq:CXYNN'l2}
\begin{split}
C_{\mathcal{N}+}&=1+\frac{\weight(m+2)P_{\mathcal{N}}}{\tilde{c}r\wt{H}},\quad C_{\mathcal{N}-}=b_-\Bigl(1+\frac{6a_-\weight(m+2)x}{r\wt{H}}\Bigr)\\
C_{\mathcal{N}'+}&=\frac{2\weight^2P_{\mathcal{N}}}{\tilde{c}\wt{H}},\quad C_{\mathcal{N}'-}=\frac{4b_-\weight^2(\tilde{c}-4rz)}{r\tilde{H}}
\end{split}
\end{equation}

where $\wt{H}(\sigma)=H(k^2\weight'-4r\sigma^2)$ and $P_X, P_Y, P_{\mathcal{N}}$ are smooth on $\hX$

\begin{equation}\label{eq:eqforP_XYNl2}
\begin{split}
\end{split}
\end{equation}

Since $\weight'=2 r^{-2}(\Bm-r^{-1}\BQ^2)>0$ in the  region $r\geq \ehRN>\Bm>\Bm^{-1}\BQ^2$ and thus $\wt{H}(0)=Hk^2\weight'>0$ there as well, we conclude that $\Psi_{\pm}(\sigma)$ exist down to $\sigma=0$. We define $\Psi_{\pm}:=\Psi_{\pm}(0)$. Conversely, using the first two equations in (<ref>) with $\sigma=0$, one can recover $X, Y$ in terms of $\Phi, \mathcal{N},\mathcal{N}'$ and thus $\Psi_{\pm}, \mathcal{N},\mathcal{N}'$. One can also check that $\Psi_{\pm}$ satisfy the equation (<ref>) with $\sigma=0$ (because $\Psi_{\pm}(\sigma)$ is holomorphic near $\sigma=0$), i.e.,

\begin{equation}\label{eq:eqforPsi+-sigma0l2}
\end{equation}

with $V_{\pm}$ defined as before in (<ref>), (<ref>), (<ref>) and (<ref>).

Since $(\dg, \dA)\in\eHb^{\infty, \ell}(\CX; \scsym\oplus\scform)\subset \mathcal{A}^{\ell+3/2}$, we have $X, Y,\mathcal{N}'\in\mathcal{A}^{\ell+3/2}(\CX)$ and $\mathcal{N}\in \mathcal{A}^{\ell+1/2}$. Let $\rho=1/r$, we notice that $H=m+6\Bm\rho-4\BQ^2\rho^2$ and $k^2\weight'=2k^2\Bm\rho^2-2k^2\BQ^2\rho^3$, so $\wt{H}(0)\in 2k^2m\Bm\rho^2+\rho^3C^\infty(\CX)$. In view of the expression in (<ref>), we find $P_Y\in C^\infty(\CX)$ and $P_X, P_{\mathcal{N}}\in \rho C^\infty(\CX)$, and then $C_{Y\pm}\in \rho^{-2}C^\infty(\CX)$, $C_{X\pm},C_{\mathcal{N}'\pm}\in \rho^{-1}C^\infty(\CX)$ and $C_{\mathcal{N}\pm}\in C^\infty(\CX)$. As a consequence, we have $\Psi_{\pm}\in\mathcal{A}^{\ell-1/2}\subset \mathcal{A}^{-2+}$ a priori because $-3/2<\ell<-1/2$. Now we are ready to describe the asymptotic behavior of $\Psi_\pm$ near $r=\ehRN$ and $r=\infty$. We only present the proof of $\Psi_+$ because the proof of $\Psi_-$ follows analogously. Since $F_{\Phi}, V_{\mathcal{N}}, a_{\pm}, b_{\pm}$ are smooth across the event horizon, we have $\weight(\pa_r\Psi_+)\overline{\Psi}_+|_{r=\ehRN}=0$. This implies that boundary term at $r=\ehRN$ arising in the integration by parts we will do later on vanishes. Near $r=\infty$, owing to (<ref>), (<ref>) (<ref>) and (<ref>), we conclude $F_{\Phi}\in\rho^3C^\infty(\CX)$, $V_{\mathcal{N}}\in \frac{k^2}{r^2}+\rho^4C^\infty(\CX)$ and then $V_{\pm}\in\frac{k^2}{r^2}+\rho^3C^\infty(\CX)$, which implies $V_{\pm}=\frac{k^2}{r^2_*}+\mathcal{A}^{3-}$ as a function of $r_*$ near $r_*=\infty$. We set $x=1/r_*$ for the moment,
\[
\]

which implies $x=0$(i.e. $r_*=\infty$) is a regular singular point and its indicial equation is $\lambda(\lambda-1) +2\lambda-k^2=0$. The roots to the indicial equation are $\lambda_{\pm}=(-1\pm\sqrt{1+4k^2})/2$. Since $k^2\geq 6$, we see that $\lambda_+\geq 2$  corresponds to one solution $\Psi_+$ with decay $\sim r_*^{-\lambda_+}$ near $r_*=\infty$, while $\lambda_-\leq -3$ implies another linearly independent solution $\Psi_+$ with growth $\sim r_*^{-\lambda_-}$ near $r_*=\infty$. Since we have $\Psi_+\in \mathcal{A}^{-2+}$ a priori, this excludes the second solution with growth. Namely, $\Psi_+\sim r^{-\lambda_+}$ which means that $\Psi_+\sim r_*^{-2}$ near $r_*=\infty$. With this decay rate near $r_*=\infty$, we can justify the $L^2(\BR_{r_*};dr_*)$ pairing between $((\weight\pa_r)^2-V_+)\Psi_+$ and $\overline{\Psi}_+$ and also find that the boundary term at $r_*=\infty$ vanishes when integrating by parts. Therefore
we proceed in the same way as in the proof of the case $\IM \sigma>0$ (use the $L^2(\BR_{r_*}; dr_*)$ pairing and then integrate by parts) and conclude that $\Psi_{\pm}=0$ on $(\ehRN, \infty)$. Therefore, $(\dg, \dA)$ is a pure gauge solution.

§.§.§ Modes with $l=1$

We now let $k^2=l(l+1)=2$. Our goal is again to prove that an outgoing mode solution $(\dg, \dA)$ is a pure gauge solution, with the gauge potential an outgoing mode as well. The scalar perturbations with $l=1$ take the form

\begin{equation}\label{eq:perturbation_l1}
\dg=\begin{pmatrix}
\widetilde{f}\sfS\\
\end{pmatrix},\quad \dA=\begin{pmatrix}
\widetilde{K}\sfS\\
\end{pmatrix}\quad\mbox{with}\quad \sfS\in\BFS_l ~(l=1).
\end{equation}

where the term $H_T$ is absent because $\rsH_1\sfS=0$. Now we follow the notations and expressions in the case $l\geq 2$ with $H_T$ set to be $0$. First, the pure gauge solutions take the form

\begin{equation}
\bdelta\dg=2\delta_g^*\omega=\begin{pmatrix}
2\hdelta^*T	\\
-\frac{r}{k}(-\frac{k}{r}T+r\hd r^{-1}L)\sd\sfS\\
\end{pmatrix},\quad\!\! \bdelta\dA=\mathcal{L}_{\omega^\sharp}A+d\phi=\begin{pmatrix}
(\hd\iota_AT+\iota_{T}\hd A+\hd P)\sfS\\
\end{pmatrix}
\end{equation}

\begin{equation}
\omega=\begin{pmatrix}
\end{pmatrix},\quad\!\! \phi=P\sfS
\end{equation}

where $T\in C^\infty(\hX; T^*\hX), L, P\in C^\infty({\hX})$.  When adding $(\bdelta\dg, \bdelta\dA)$ to $(\dg, \dA)$, the quantities $\tilde{f}, f, H_L, \tilde{K}, K$ change by

\begin{equation}
\begin{gathered}
\bdelta\widetilde{f}=2\hdelta^*T,\quad\bdelta f=-\frac{k}{r}T+r\hd (r^{-1}L),\quad \bdelta H_L=r^{-1}\iota_{dr}T+\frac{k}{2r}L,\quad %\bdelta H_T=-\frac{k}{r}L;\\
\\
\bdelta\widetilde{K}=\hd\iota_AT+\iota_T\hd A+\hd P,\quad \bdelta K=-\frac{k}{r}(\iota_A T+P).
\end{gathered}
\end{equation}

We define $\mathbf{X}:=\frac{r}{k}f$, then $\bdelta\mathbf{X}=\frac{r}{k}\bdelta f=-T+\frac{r^2}{k}\hd(r^{-1}L)$. We observe that for any given $L$, one can choose

\begin{equation}\label{eq:puregaugen=1}
T=\mathbf{X}+\frac{r^2}{k}\hd(r^{-1}L),\quad P=-\iota_AT+\frac{r}{k}K
\end{equation}

which implies $\mathbf{X}+\bdelta\mathbf{X}=0, K+\bdelta K=0$.  If $(\dg, \dA)$ are outgoing mode solutions and $L$ is a mode, then $T, P$ are modes of the same type which grows at most by the factor $r$ relative to $\dg, \dA$ and $L$.  We again define

\begin{equation}
\begin{split}
\widetilde{F}&:=\widetilde{f}+2\hdelta^*\mathbf{X}\in C^{\infty}(\hX; S^2T^*\hX),\\
J&:=H_L+r^{-1}\iota_{dr}\mathbf{X}\in C^\infty(\hX;T^*\hX),\\
N&:=\widetilde{K}+\hd\left(\frac{r}{k}K\right)+\iota_{\mathbf{X}}\hd A\in C^\infty(\hX; T^*\hX).	\end{split}
\end{equation}

However, $\wt{F}, J, N$ are not gauge-invariant in this case. In fact

\begin{equation}
\bdelta\wt{F}=\frac{2}{k}\hdelta^*\Bigl(r^2\hd(r^{-1}L)\Bigr),\quad\bdelta J=\frac{k}{2r}L+\frac{r}{k}\iota_{dr}\hd(r^{-1}L),\quad \bdelta N=\frac{r^2}{k}\iota_{\hd(r^{-1}L)}\hd A.
\end{equation}

For any fixed $L$, we pick $T, P$ as explained above and then can work in the gauge $\mathbf{X}=0, K=0$, which implies

\[
(\wt{f}, f,H_L, \wt{K}, K)=(\wt{F},0,J,N,0),
\]

and thus the linearized Einstein-Maxwell system $2\mathcal{L}_1(\dg, \dA)=0, \mathcal{L}_2(\dg, \dA)=0$ is again given by (<ref>)–(<ref>) with the equation (<ref>) for $H^E_T$ absent. Conversely, in the gauge $\mathbf{X}=0, K=0$, if $(\wt{F}, J, N)=0$, we have $(\dg, \dA)=0$, which implies the original perturbations $(\dg, \dA)$ take the form

\begin{equation}\label{eq:gaugesolnl=1}
(\dg, \dA)=(2\delta_g^*\omega, \mathcal{L}_{\omega^\sharp}A+d\phi) \quad\!\! \mbox{with}\quad\!\! \omega=-\begin{pmatrix}
\Bigl(\mathbf{X}+\frac{r^2}{k}\hd(r^{-1}L)\Bigr)\sfS\\
\end{pmatrix},\quad \!\! \phi=\iota_A(\mathbf{X}+\frac{r^2}{k}\hd(r^{-1}L))-\frac{r}{k}K.
\end{equation}

If $(\dg, \dA)$ are outgoing mode solutions and $L$ is a mode , then $\omega$ and $ \phi$ are modes of the same type which grow at most by a factor $r$ more than $(\dg, \dA)$ and $L$.

Now we exploit this extra gauge freedom resulting from $L$ to simplify the structure of the linearized Einstein-Maxwell system. Concretely, we define $H_T^E:=-\frac{k^2}{2r^2}\htr\wt{F}$ and we shall arrange $H^E_T+\bdelta H_T^E=0$ by choosing $L$ properly and thus the absent equation (<ref>) still holds in the case $l=1$. In order to achieve this, we should choose $L$ such that

\begin{equation}\label{eq:eqnforL}
\begin{split}
\bdelta H_T^E&=-\frac{k}{r^2}\htr\hdelta^*\Bigl(r^2\hd(r^{-1}L)\Bigr)=\frac{k}{r^2}\hdelta\Bigl(r^2\hd(r^{-1}L)\Bigr)=-k\hBox_{\hg}(r^{-1}L)-2kr^{-1}\pa^i(r)\pa_i(r^{-1}L)\\
\end{split}
\end{equation}

We first consider the non-stationary mode solutions with $\IM \sigma\geq0,  \sigma\neq0$. We again define $\mathcal{N}$ as $\hstar N=\hd\mathcal{N}$ and $X, Y,Z$ as in (<ref>). As explained in the case $l\geq 2$, $\mathcal{N}, X, Y,Z$ are modes with the same frequency $\sigma$. We shall find a mode $L$ satisfying the above equation (<ref>). Writing $H_T^E=e^{-i\sigma t_*}H_{T,0}^E(r)$, then the equation (<ref>) for $L=e^{-i\sigma t_*}\hat{L}(r)$ becomes

\begin{equation}
\widehat{\Box_g}(\sigma)(r^{-1}\hat{L})=\frac{H_{T,0}^E}{k}
\end{equation}

which has a spherically symmetric solution since by Theorem <ref> $\widehat{\Box_g}(\sigma): \eHb^{\infty, \ell}(\CX)\to\eHb^{\infty, \ell+1}(\CX)$ is invertible for any $\ell<-1/2$. Therefore, as discussed around (<ref>), we add a pure gauge solution using this $L$ and the corresponding $T, P$ as in (<ref>) , the linearized Einstein-Maxwell system is again given by (<ref>)–(<ref>). This allows us to repeat the proof of the case of non-stationary mode solutions with $l\geq 2$ to conclude that in the case $l=1$, $(\dg, \dA)$ is a gauge solution as well. More precisely, $(\dg, \dA)$ take the form as in (<ref>).

Next we discuss the stationary case $\sigma=0$. Suppose $(\dg, \dA)\in\eHb^{\infty, \ell}(\CX; \scsym\oplus\scform)$ where we now assume $-5/2<\ell<-1/2$, then $\mathbf{X}=rf/k\in \eHb^{\infty, \ell-1}(\CX; \scform)$ and thus $\hdelta^*\mathbf{X}\in\eHb^{\infty, \ell}(\CX; \scsym)$ (because $\hdelta^*$ acts as $\rho\mbox{Diff}^1_b$). Therefore, the right-handed side of (<ref>) is in $\eHb^{\infty, \ell+2}(\CX)$. According to Theorem <ref>,
$\widehat{\Box_g}(0): \eHb^{\infty, \ell}(\CX)\to \eHb^{\infty, \ell+2}(\CX)$ is surjective for $-5/2<\ell<-1/2$,
and then we can solve (<ref>) for $r^{-1}L\in \eHb^{\infty, \ell}(\CX)$ and thus $L\in \eHb^{\infty, \ell-1}(\CX)$. Again, according to (<ref>), by adding a pure gauge solution $(2\hdelta_g^*\omega, \mathcal{L}_{\omega^\sharp}A+d\phi)$ with $(\omega, \phi)\in \eHb^{\infty, \ell-1}(\CX; \scform\oplus\BR)$, the linearized Einstein-Maxwell system is again given by (<ref>)–(<ref>).

Next, we need to modify the argument in the case $l\geq 2, \sigma=0$ a little bit. We again write

\begin{equation}\label{eq:newexpressionforPsi+-l=1}
\Psi_{\pm}(\sigma)=C_{X\pm}(\sigma)X+C_{Y\pm}(\sigma)Y+C_{\mathcal{N}\pm}(\sigma)\mathcal{N}+C_{\mathcal{N}'\pm}(\sigma)\mathcal{N}'
\end{equation}

where $C_{\bullet \pm}(
\sigma)$ are rational functions of $r$ depending on $\sigma\in\BC$ (obtained by letting $k^2=2, m=0$ in (<ref>))

\begin{equation}\label{eq:CXYNN'l1}
\begin{split}
C_{\mathcal{N}+}&=1+\frac{2\weight P_{\mathcal{N}}}{\tilde{c}r\wt{H}},\quad C_{\mathcal{N}-}=b_-\Bigl(1+\frac{12a_-\weight x}{r\wt{H}}\Bigr)\\
C_{\mathcal{N}'+}&=\frac{2\weight^2P_{\mathcal{N}}}{\tilde{c}\wt{H}},\quad C_{\mathcal{N}'-}=\frac{4b_-\weight^2(\tilde{c}-4rz)}{r\tilde{H}}
\end{split}
\end{equation}

where $\wt{H}(\sigma)=H(2\weight'-4r\sigma^2)$ with $H(r)=6\Bm r^{-1}-4\BQ^2 r^{-2}$, $(a_+, b_+)=(\frac{\BQ}{2r}, 1)$, $(a_-, b_-)=(1-\frac{2\BQ^2}{3\Bm r}, -\frac{4\BQ}{3\Bm})$, $\tilde{c}=6\Bm$ and $P_X, P_Y, P_{\mathcal{N}}$ are smooth on $\hX$(we again let $k^2=2, m=0$ in (<ref>))

\begin{equation}\label{eq:eqforP_XYNl1}
\begin{split}
\end{split}
\end{equation}

Since $H(r)=2r^{-1}(3\Bm-2\BQ^2 r^{-1})>0$ and $\weight'=2 r^{-2}(\Bm-r^{-1}\BQ^2)>0$ in the  region $r\geq\ehRN>\Bm>\Bm^{-1}\BQ^2>(2/3)\Bm^{-1}\BQ^2$, we have $\wt{H}(0)=2H\weight'>0$ there as well. We again define $\Psi_{\pm}:=\Psi_{\pm}(0)$. Conversely, using the the first two equations in (<ref>) with $\sigma=0$ and $m=0$, one can recover $X, Y$ in terms of $\Phi, \mathcal{N},\mathcal{N}'$ and thus $\Psi_{\pm}, \mathcal{N},\mathcal{N}'$ as well. Again, $\Psi_{\pm}(0)$ satisfy

\begin{equation}\label{eq:eqforPsi+-sigma0l1}
\end{equation}

with $V_{\pm}$ defined as before in (<ref>) (<ref>), (<ref>) and (<ref>) with $m=0, k^2=2$.

Since $(\dg, \dA)\in\eHb^{\infty, \ell}(\CX; \scsym\oplus\scform)\subset \mathcal{A}^{\ell+3/2}(\CX)$, we have $X, Y, \mathcal{N}'\in\mathcal{A}^{\ell+3/2}(\CX)$ and $\mathcal{N}\in \mathcal{A}^{\ell+1/2}(\CX)$. Let $\rho=1/r$, we notice that $H=6\Bm\rho-4\BQ^2\rho^2$ and $k^2\weight'=4\Bm\rho^2-4\BQ^2\rho^3$, so $\wt{H}(0)\in 24\Bm^2\rho^3+\rho^4C^\infty(\CX)$. In view of the expression in (<ref>), we find $P_X,  P_Y\in \rho C^\infty(\CX)$ and $P_{\mathcal{N}}\in \rho^2 C^\infty(\CX)$, and then $C_{X+}, C_{Y+}, C_{\mathcal{N}'+}\in \rho^{-1}C^\infty(\CX)$ and $C_{\mathcal{N}+}\in C^\infty(\CX)$. As a consequence, we have

\[
\Psi_{+}\in\mathcal{A}^{\ell+1/2}(\CX)\subset \mathcal{A}^{-2+}(\CX)\quad\!\!\mbox{a priori if}\!\!\quad -5/2<\ell<-1/2.
\]

Now we are ready to describe the asymptotic behavior of $\Psi_+$ near $r=\ehRN$ and $r=\infty$. Since $F_{\Phi}, V_{\mathcal{N}}, a_{\pm}, b_{\pm}$ are smooth across the event horizon, we have $\weight(\pa_r\Psi_+)\overline{\Psi}_+|_{r=\ehRN}=0$. This implies that boundary term at $r=\ehRN$ arising in the integration by parts we will do later on vanishes. Near $r=\infty$, owing to (<ref>), (<ref>) (<ref>) and (<ref>) with $m=0$, we conclude $F_{\Phi}\in \frac{8\BQ}{3\Bm}\rho^2+\rho^3C^\infty(\CX)$, $V_{\mathcal{N}}\in \frac{2}{r^2}+\rho^3C^\infty(\CX)$ and then $V_{+}\in\frac{2}{r^2}+\rho^3C^\infty(\CX)$, which implies $V_{+}=\frac{2}{r^2_*}+\mathcal{A}^{3-}$ as a function of $r_*$ near $r_*=\infty$. We set $x=1/r_*$ for the moment,
\[
\]

which implies $x=0$(i.e. $r_*=\infty$) is a regular singular point and its indicial equation is $\lambda(\lambda-1) +2\lambda-2=0$. The roots to the indicial equation are $\lambda_+=1, \lambda_-=-2$. Again we see that  $\lambda_+=1$  corresponds to one solution $\Psi_+$ with decay $\sim r_*^{-1}$ near $r_*=\infty$, while $\lambda_-\leq -2$ implies another linearly independent solution $\Psi_+$ with growth $\sim r_*^{2}$ near $r_*=\infty$. Since we have $\Psi_+\in \mathcal{A}^{-2+}$ a priori, this excludes the second solution with growth. Namely,  $\Psi_+\sim r_*^{-1}$ near $r_*=\infty$. With this decay rate near $r_*=\infty$, we can justify the $L^2(\BR_{r_*};dr_*)$ pairing between $((\weight\pa_r)^2-V_+)\Psi_+$ and $\overline{\Psi}_+$ and also find that the boundary term at $r_*=\infty$ vanishes when integrating by parts. Therefore
we proceed in the same way as in the proof of the case $\IM \sigma>0$ (use the $L^2(\BR_{r_*}; dr_*)$ pairing and then integrate by parts) and conclude that $\Psi_+=0$ on $(\ehRN, \infty)$ and thus on $\hX$.

As for $\Psi_-$, recall the S-deformation we discussed around (<ref>) and we can rewrite the equation $((\weight\pa_r)^2-V_-)\Psi_-=0$ as

\[
\wt{D}^*\wt{D}\Psi_-=-H_-\pa_r\Bigl(H_-^{-2}\weight\pa_r(H_-\Psi_-)\Bigr)=0\quad \mbox{with}\quad H_-(r)=\frac{\tilde{c}}{r}=\frac{6\Bm}{r}>0\ \mbox{on}\ r\in[\ehRN, \infty)
\]

because $\wt{V}_-=\frac{k^2(k^2-2)\weight}{r^2H_-}=0$ if $l=1$. Solving the above equation for $\Psi_-$ yields

\begin{equation}\label{eq:SPsi_-}
\Psi_-=\frac{c_1}{H_-}\int\frac{H_-^2}{\weight}dr+\frac{c_2}{H_-}=d_1\ln\left\lvert\frac{r-r_H}{r-(\Bm-\sqrt{\Bm^2-\BQ^2})}\right\rvert+d_2r.
\end{equation}

where $c_1, c_2, d_1, d_2\in \BR$. Since $\Psi_-$ is smooth across the event horizon $r=\ehRN$, we conclude $d_1=0$ and thus $\Psi_-=d_2r$. Using $(\Psi_+, \Psi_-)=(0, d_2 r)$,  (<ref>) and (<ref>), we have $(\Psi, \mathcal{N})=(d_2r, -d_2\BQ/2)$. Exploiting the first two equations in (<ref>) we obtain $(X, Y)=(-d_2,-d_2)$. As discussed previously, the equation (<ref>) implies $Z=0$ in the stationary perturbation. Finally using (<ref>) and $\hstar N=\hd\mathcal{N}$ we see

\[
\wt{F}=0, \quad J=-\frac{d_2}{2},\quad  N=0
\]

Since $J\in\eHb^{\infty, \ell}(\CX)$ and $0\neq d_2\in\eHb^{\infty, -\frac{3}{2}-}(\CX)$, we have $d_2=0$ if $\ell\geq-3/2$ and thus we are done. If $-5/2<\ell<-3/2$, $d_2$ may be non-zero and then in the gauge $\mathbf{X}=0$

\[
\dg=\begin{pmatrix}
\end{pmatrix}, \quad \dA=0.
\]

This implies in the gauge $\mathbf{X}=0$, $(\dg, \dA)$ is a gauge solution with $T=0, L=-d_2r/k, P=0$. Therefore, according to the discussion around (<ref>), the original perturbations $(\dg, \dA)=(2\delta_g^*\omega, \mathcal{L}_{\omega^\sharp}A+d\phi)$ with $(\omega, \phi)\in \eHb^{\infty, \ell-1}(\CX; \scform\oplus\BR)$ for $-5/2<\ell<-1/2$.

§.§.§ Modes with $l=0$

As stated in [60, 57], the linearization of the simple proof of the Birkhoff' theorem [103] can be extended to the spherically symmetric Einstein-Maxwell system. Following the argument in [103], we work in the incoming Eddington-Finkelstein coordinates $(t_0=t+r^*, r)$ where the Reissner-Nordström metric and the electromagnetic $4$-potential take the form

\begin{equation}\label{eq:incomingEF}
g=-\weight dt_0^2+2dt_0dr+r^2\sg,\quad A=\frac{\BQ}{r}dt_0
\end{equation}

The electromagnetic $4$-potential defined above differs from (<ref>) by an exact $1$-form $f(r)dr$. We note that adding an exact $1$-form to $A$ changes neither the electromagnetic field nor the linearized Einstein-Maxwell system because the linearized system only involves $dA$.

Representing the Reissner-Nordström metric and the electromagnetic $4$-potential in the way (<ref>), we can linearize $(g_b,A_b)$ with respect to the parameters $(\Bm, 0,\BQ)$ in the direction $(\dot{\Bm},0, \dot{\BQ})$ to obtain the following linearized metric and  electromagnetic $4$-potential

\begin{equation}\label{eq:linearizedRNv}
g'=(\frac{2\dot{\Bm}}{r}-\frac{2\BQ\dot{\BQ}}{r^2})dt_0^2,\quad A'=\frac{\dot{\BQ}}{r}dt_0.
\end{equation}

We note that $(g', A')$ differs from $(\dg_{(\Bm, 0, \BQ)}(\dot{\Bm}, 0, \dot{\BQ}), \dA_{(\Bm, 0, \BQ)}(\dot{\Bm}, 0, \dot{\BQ}))$ by a pure gauge term.

With this we now turn to the analysis of the spherically symmetric modes. Since there is no spherical component in $1$-forms and no aspherical-spherical mixed component in symmetric $2$-tensors for spherically symmetric perturbations, we use the following splitting of $1$-forms and symmetric $2$-tensors respectively:$ \langle dt_0\rangle\oplus \langle dr\rangle$ and $\langle dt_0^2\rangle\oplus \langle 2dt_0dr\rangle\oplus\langle dr^2\rangle\oplus \langle\sg\rangle$. Under this splitting,  we write the scalar perturbations of modes $l=0$ (spherically symmetric modes) in the following form

\[
\dg=\begin{pmatrix}
\dot{X}\\
\dot{Y}\\
\dot{Z}\\
\end{pmatrix},\quad \dA=\begin{pmatrix}
\dA_0\\
\dA_1
\end{pmatrix}
\]

where $\dot{X}, \dot{Y}, \dot{Z}, H_L$ are smooth function on $\BR_{t_0}\times [r_-, \infty)$. To describe the pure gauge perturbations, we use the following lemma

Let $(g, A)	$ be as defined in (<ref>) , under the splitting $\langle\pa_{t_0}\rangle\oplus\langle\pa_r\rangle$ of $TM$ and $\langle dv^2\rangle\oplus \langle 2dt_0dr\rangle\oplus\langle dr^2\rangle$ of $S^2T^*M$, the map $\mathcal{L}_{(\bullet)}g: TM\to S^2T^*M$ takes the form

\begin{equation}
\mathcal{L}_{(\bullet)}g=\begin{pmatrix}
\pa_{t_0}-\weight\pa_r&\pa_r\\
\end{pmatrix};
\end{equation}

the map $\mathcal{L}_{(\bullet)}A$ between sections of $\langle \pa_{t_0}\rangle\oplus \langle \pa_r\rangle$ and $\langle dt_0\rangle\oplus \langle dr\rangle$ is given by

\begin{equation}
\mathcal{L}_{(\bullet)}A=\begin{pmatrix}
\BQ r^{-1}\pa_{t_0}&-\BQ r^{-2}\\
\BQ r^{-1}\pa_r& 0
\end{pmatrix},
\end{equation}

and  as a mapping between the sections of the trivial line bundle $\BR$ and $\langle dt_0\rangle\oplus \langle dr\rangle$ , $d=\begin{pmatrix}
\pa_{t_0}\\
\pa_r
\end{pmatrix}$.

It follows from the formula $(\mathcal{L}_Wg)_{\alpha\beta}=W^\gamma\pa_\gamma g_{\al\be}+(\pa_{\al}W^\gamma)g_{\gamma\be}+(\pa_\be 	W^\gamma)g_{\al\gamma}$, $(\mathcal{L}_W A)_\al=W^\gamma\pa_\gamma A_{\al}+(\pa_{\al}W^\gamma)A_{\gamma}$.

Therefore, adding a pure gauge solution $(\mathcal{L}_Wg, \mathcal{L}_W A)$ to $(\dg, \dA)$ with $W:=Z_1\pa_{t_0}-rH_L\pa_r$ and $Z_1:=(-\int_{3\Bm}^r \dot{Z}(t_0, s)ds)/2$, the $dr^2$ and spherical components of $\dg$ vanish. Next, by adding another pure gauge solution $(0, d\phi)$ with $\phi=-\int_{3\Bm}^r(A_1 +\BQ s^{-1}\pa_r Z_1)(t_0, s)ds$, the $dr$-component of $\dA$ can be eliminated.

Suppose the original perturbations $(\dg, \dA)\in e^{-i\sigma t_*}\eHb^{\infty, \ell}(\CX;\scsym\oplus\scform)$ are outgoing modes, then so are $\dot{X}, \dot{Y}, \dot{Z}, H_L$. If $\sigma=0$, it is obvious that $(\omega=W^\flat, \phi)\in \eHb^{\infty, \ell-1}(\CM; \scform\oplus \BR)$ are still stationary modes. If $\sigma\neq 0$, since $t_*=t_0+f(r)$ where $f$ is smooth on $[r_-, \infty)$ and $\sim -2 r^*$ near infinity, we find that $(\omega=W^\flat, \phi)\in e^{-i\sigma t_*}\eHb^{\infty, \ell'}(\CX;\scform\oplus\BR)$ for some $\ell'<\ell$ is also outgoing modes. Further, if $(\dg, \dA)\in \mbox{Poly}(t_*)^k\eHb^{\infty, \ell}(\CX;\scsym\oplus\scform)$ are the generalized stationary modes, the integration in the definition of $W$ and $\phi$ implies $(\omega=W^\flat, \phi)\in \mbox{Poly}(t_*)^k\eHb^{\infty, \ell'}(\CX;\scform\oplus\BR)$ for some $\ell'<\ell$. To summarize, by adding pure gauge solutions which are outgoing modes(or generalized stationary modes), the perturbations are reduced to the following form

\[
\dg=\dot{X}dt_0^2+2\dot{Y}dt_0dr,\quad \dA=\dA_0dt_0.
\]

Next, we analyze the linearized Einstein-Maxwell system for spherically symmetric modes. Concretely, for $(\check{g}, \check{A})$ of the general form $\check{g}=Xdt_0^2+2Ydt_0dr+r^2\sg$ and $\check{A}=A_0dt_0$ where $X, Y,A_0$ are functions of $(t_0,r)$, the $dr^2$-component of $\Ric(\check{g})-2T(\check{g}, d\check{A})=0$ implies $2\pa_r Y/(rY)=0$. Linearizing it around $(g, A)$ in (<ref>), we obtain the corresponding equation for $(\dg, \dA)$: $\pa_r \dot{Y}=0$, so $\dot{Y}=\dot{Y}(t_0)$. If $\dot{Y}$ is an outgoing mode with $\sigma\neq 0$, we must have $\dot{Y}=\dot{Y}(t_0)=ce^{-i\sigma t_0}$ which contradicts the definition of outgoing modes unless $\dot{Y}=0$. If $\dot{Y}$ is a stationary mode, then $\dot{Y}=c$. However the assumption $\dot{Y}\in \eHb^{\infty, \ell}(\CX)\subset \mathcal{A}^{0+}(\CX)$ for $-3/2<\ell<-1/2$ require $\dot{Y}=0$. Lastly, if $\dot{Y}$ is a generalized zero mode, we then have $\dot{Y}=\dot{Y}(t_0)=\mbox{Poly}(t_0)^k$. Letting $W_1=f\pa_{t_0}$ with $f(t_0)\in\mbox{Poly}(t_0)^{k+1}$ satisfying $\pa_{t_0} f(t_0)=-\dot{Y}(t_0)$, and adding the pure gauge solution $(\mathcal{L}_{W_1}g, \mathcal{L}_{W_1}A=0)$ to $(\dg, \dA)$, $dvdr$-component of $\dg$ again vanishes. Therefore, we now have

\begin{equation}
\dg=\dot{X}dt_0^2,\quad \dA=\dA_0dt_0.
\end{equation}

Again, for $(\check{g}, \check{A})$ the Maxwell equation becomes

\begin{equation}
\begin{split}
\delta_{\check{g}}d\check{A}&=\Bigl(-\frac{1}{Y}\pa_{t_0}\pa_r A_0+\frac{X}{Y^2}\pa^2_r A_0+\frac{2X}{rY^2}\pa_r A_0-\frac{X}{Y^3}\pa_r Y\pa_r A_0+\frac{1}{Y^2}\pa_r A_0\pa_{t_0} Y\Bigr)dv\\
&\quad+\Bigl(\frac{1}{Y}\pa_r^2 A_0+\frac{2}{rY}\pa_r A_0-\frac{1}{Y^2}\pa_r Y\pa_rA_0\Bigr) dr.
\end{split}
\end{equation}

Linearizing around $(g, A)$ (i.e. $X=-\weight, Y=1, A_0=r^{-1}\BQ$) and using the previously obtained fact $\dot{Y}=0$, the $dr$ component tells $\pa_r^2\dA_0+2r^{-1}\pa_r\dA_0=0$, so $\dA_0=c_1(t_0)r^{-1}+c_2(t_0)$. Since $c_2(t_0)dt_0$ is an exact $1$-form and thus a pure gauge solution, we may assume $\dA_0=c_1(t_0)r^{-1}$. Then the equation for $dt_0$-component gives $-\pa_{t_0}\pa_r\dA_0-\weight\pa_r^2\dA_0-2\weight r^{-1}\pa_r\dA_0=r^{-2}\pa_{t_0}c_1(t_0)=0$ and thus $c_1(t_0)$ is a constant that we denote by $\dot{\BQ}$. That is, $\dA_0=r^{-1}\dot{\BQ}$.

Next, the spherical component of $\Ric(\check{g})-2T(\check{g}, d\check{A})=0$ is given by

\[
1+\frac{X}{Y^2}+\frac{r}{Y^2}\pa_r X-\frac{rX}{Y^3}\pa_r Y-\frac{r^2}{Y^2}(\pa_r A_0)^2=0.
\]

Linearizing around $(g, A)$ (i.e. $X=-\weight, Y=1, A_0=r^{-1}\BQ$) and using the previously obtained facts $\dot{Y}=0, \dA_0=r^{-1}\dot{\BQ}$, we find $\dot{X}+r\pa_r \dot{X}-2r^{-2}\BQ\dot{\BQ}=0$ and thus

\[
\dot{X}=-\frac{2\BQ\dot{\BQ}}{r^2}+\frac{c(t_0)}{r}.
\]

Lastly, the $dt_0^2$-component of  $\Ric(\check{g})-2T(\check{g}, d\check{A})=0$ is

\[
-\frac{X}{Y^2}\pa_{t_0}\pa_r Y+\frac{X}{2Y^2}\pa_r^2 X+\frac{X}{Y^3}\pa_{t_0}Y\pa_r Y-\frac{X}{2Y^3}\pa_r X\pa_r Y-\frac{2X}{rY^2}\pa_{t_0} Y+\frac{X}{rY^2}\pa_r X+\frac{1}{rY}\pa_{t_0} X+\frac{X}{Y^2}(\pa_r A_0)^2=0.
\]

Again linearizing around $(g, A)$ (i.e. $X=-\weight, Y=1, A_0=r^{-1}\BQ$) and using the previously obtained facts $\dot{Y}=0, \dA_0=r^{-1}\dot{\BQ}, \dot{X}=-\frac{2\BQ\dot{\BQ}}{r^2}+\frac{c(v)}{r}$, we obtain $r^{-2}\pa_vc(v)=0$ and thus $c(v)$ is a constant which we denote by $2\dot{\Bm}$. Therefore, we conclude that up to pure gauge solutions

\begin{equation}
\dg=(\frac{2\dot{\Bm}}{r}-\frac{2\BQ\dot{\BQ}}{r^2})dt_0^2=g',\quad \dA_0=\frac{\dot{\BQ}}{r}dt_0=A'.
\end{equation}

§.§ Vector type perturbations

We shall consider the vector type perturbations of modes $l\geq 2$ and $l=1$ separately.  Recall that we can write the perturbations under the splitting (<ref>) and (<ref>) as in (<ref>). We denote by $\sfV\in\BFV_l$ the spherical harmonic $1$-form with eigenvalues $k^2=l(l+1)-1$.

§.§.§ Modes with $l\geq 2$

Suppose $(\dg, \dA)$ is the vector perturbations of the following form

\begin{equation}\label{eq:vperturbation_l2}
\dg=\begin{pmatrix}
\end{pmatrix},\quad \dA=\begin{pmatrix}
\end{pmatrix}\quad\mbox{with}\quad \sfV\in\BFV_l ~(l\geq2).
\end{equation}

We notice that the pure gauge solutions take the form

\begin{equation}
\bdelta\dg=2\delta_g^*\omega=\begin{pmatrix}
0	\\
\end{pmatrix},\quad\!\! \bdelta\dA=\mathcal{L}_{\omega^\sharp}A=0
\end{equation}

\begin{equation}
\omega=\begin{pmatrix}
\end{pmatrix}
\end{equation}

where $L\in C^\infty({\hX})$. When adding $(\bdelta\dg, \bdelta\dA)$ to $(\dg, \dA)$, the quantities $ f,  H_T, K$ change by

\begin{equation}
\bdelta f=r\hd (r^{-1}L),\quad \bdelta H_T=-kr^{-1}L, \quad \bdelta K=0.
\end{equation}

Then we have the following gauge-invariant quantities

\begin{equation}
J:=f+\frac{r}{k}\hd H_T\in C^\infty(\hX; T^*\hX),\quad K\in C^\infty(\hX)
\end{equation}

We also note that if $J=0, K=0$, $(\dg, \dA)$ is a pure gauge solution

\begin{equation}
(\dg, \dA)=(2\delta^*\omega, \mathcal{L}_{\omega^\sharp}A) \quad\!\! \mbox{with}\quad\!\! \omega=\begin{pmatrix}
\end{pmatrix}
\end{equation}

If $(\dg, \dA)$ are outgoing mode solutions, then $\omega$ is a mode of the same type, which grows by a factor $r$ more than $(\dg, \dA)$.

Again we can express the linearized Einstein-Maxwell system in terms of the gauge-invariant quantities $J, K$ defined above.
More specifically, we adds $(\bdelta\dg, \bdelta\dA)$ built from $ L=\frac{r}{k}H_T$, then $ H_T+\bdelta H_T=0$, and thus $(f, H_T,K)=(J, 0, K)$.
Therefore, using the detailed calculation in <ref> and <ref> we can write the linearized 	Einstein-Maxwell system, acting on the new $(\dg, \dA)$

\[
\dg=\begin{pmatrix}
\end{pmatrix}, \quad \dA=\begin{pmatrix}
\end{pmatrix}
\]
in terms of the gauge-invariant quantities $J, K$.

Now we express $2\mathcal{L}_1(\dg, \dA)=0$ and $\mathcal{L}_2(\dg, \dA)=0$ in the form of (<ref>) in terms of $f^E,  H_T^E$ and $K^E$ respectively. Then the linearized Einstein-Maxwell system reads

\begin{align}
rf^E&=r^{-2}\hdelta(r^4\hd(r^{-1}J))+\frac{k^2-1}{r}J-4\BQ r^{-2}\hstar\hd(rK)=0,\label{eq:EfV}\\
rK^E&=-\hBox(rK)+\frac{k^2+1}{r}K+\BQ \hstar\hd(r^{-1}J)=0.\label{eq:EKV}
\end{align}

By (<ref>), one adds the zero term $\hd\hdelta (rJ)$ to (<ref>) and then can obtain a wave equation for $J$ (coupled to $K$ via subprincipal terms) as $(\hdelta\hd+\hd\hdelta) J=(-\hBox
-\frac 12\weight'')J$. In conclusion, we can rewrite equation (<ref>) and (<ref>) as a principally scalar system of wave equations for $( J, K)$

\begin{equation}
-\hBox B-\mathscr{D}B=0\quad \mbox{where}\quad
\begin{pmatrix}
\end{pmatrix}
\end{equation}

where $\mathscr{D}$ is a first order stationary differential operator acting on $C^{\infty}(\hX;T^*\hX \oplus T^*\hX)$.
As discussed in the scalar perturbations, it suffices to prove $B=0$ in the static region $r>\ehRN$. To achieve this, we again work in the static coordinates $(t,r)$.

First, by (<ref>) we have $\hd\hstar(rJ)=0$ (because $\hdelta=\hstar\hd\hstar$), and then $\hstar rJ=\hd(r\Phi)$ for some function $\Phi$ on $\hX$. Applying $\hstar r^2$ on both sides of equation (<ref>) yields

\[
\hd \Bigl(r^4\hdelta(r^{-2}\hd(r\Phi))+(k^2-1)r\Phi-4\BQ rK\Bigr)=0
\]

which implies $r^4\hdelta(r^{-2}\hd(r\Phi))+(k^2-1)r\Phi-4\BQ rK=c$ for some constant $c$. Replacing $\Phi$ by $\Phi-\frac{c}{(k^2-1)r}$, we obtain $
r^4\hdelta(r^{-2}\hd(r\Phi))+(k^2-1)r\Phi-4\BQ rK=0.
$ and thus

\begin{equation}\label{eq:eqnforPhiversion1}
\end{equation}

We can also rewrite it as

\begin{equation}\label{eq:eqnforPhiVl2}
\end{equation}

Next, equation (<ref>) can be rewritten as (where we use equation (<ref>))

\begin{equation}\label{eq:eqnforrK}
\end{equation}

We note that (<ref>) and (<ref>) recover <cit.> with $\varOmega=r\Phi, \mathcal{A}=-rK, m_V=k^2-1, \kappa^2=2, q=\BQ, n=2, K=1, \bar{\tau}=0, J=0$ there. Following <cit.>, we introduce the following combinations

\begin{equation}\label{eq:defforPsi+-Vl2}
\Psi_{\pm}=a_{\pm}\Phi+b_{\pm}rK
\end{equation}

\begin{equation}\label{eq:defofab+-Vl2}
(a_+, b_+)=(\frac{\BQ(k^2-1)}{3\Bm+\sqrt{9\Bm^2+4(k^2-1)\BQ^2}}, -1),\quad (a_-, b_-)=(1, \frac{4\BQ}{3\Bm+\sqrt{9\Bm^2+4(k^2-1)\BQ^2}}).
\end{equation}

When expressed in terms of $\Psi_\pm$, equations (<ref>) and (<ref>) are transformed into two decoupled wave equations (see <cit.>)

\begin{equation}\label{eq:eqnforPsi+-Vl2}
\hBox\Psi_\pm-\frac{V_{\pm}}{\weight}\Psi_\pm=0
\end{equation}

\begin{equation}
\end{equation}

Suppose $(\dg, \dA)\in e^{-i\sigma t_*}\eHb^{ \infty,\ell}(\CX; \scsym\oplus\scform)$ are modes, then $(J, K)\in e^{-i\sigma t_*}\eHb^{ \infty,\ell-1}(\CX; \scform\oplus\BR)$ for $\sigma\neq 0$, while $(J, K)\in\eHb^{\infty,\ell}(\CX; \scform\oplus\BR)$ for $\sigma=0$. As discussed around (<ref>) and (<ref>) and using (<ref>) in the $\sigma=0$ case, we conclude $\Phi, rK\in e^{-i\sigma t_*}\eHb^{ \infty,\ell-1}(\CX)$ for $\IM\sigma\geq0$, and so are $\Psi_\pm$. Therefore in the static coordinates $(t,r)$, equations (<ref>) take the form

\[
\]

We observe that when $r>\ehRN=\Bm+\sqrt{\Bm^2-\BQ^2}$, $V_+>0$ and

\begin{align*}
\end{align*}

where we use $k^2\geq 5, \Bm>\BQ$ in the last inequality, so $V_->0$ in the static region as well. We also notice the asymptotic behavior of $V_\pm$ is $V_\pm=(k^2+1)\rho^2+\rho^3C^\infty(\CX)$ where $\rho=1/r$. Then we can proceed as in the scalar perturbations $l\geq 2$ case to conclude that $\Psi_\pm=0$ and thus $(J,K)=0$. More specifically, when $\IM\sigma>0$, we pair $(\weight\pa_r)^2\Psi_\pm-(V_\pm-\sigma^2)\Psi_\pm$ with $\overline{\Psi}_\pm$ with respect to $L^2(\BR_{r_*}; dr_*)$ and integrate by parts, then the positivity of $V_\pm$ allows us to obtain $\Psi_\pm=0$. When $\sigma\in\BR\setminus\{0\}$, we use the boundary paring argument. Lastly, when $\sigma=0$, we first obtain the a priori decay rate $\Psi_\pm\in\eHb^{ \infty,\ell-1}(\CX)\subset\mathcal{A}^{\ell+1/2}(\CX)\subset\mathcal{A}^{-1+}(\CX)$ because $-3/2<\ell<-1/2$. Next we use the Frobenius method (i.e. analyze the indicial equation) and the decay rate of $V_\pm=(k^2+1)\rho^2+\rho^3C^\infty(\CX)$ to conclude that $\Psi_\pm\sim r^{-2}$ near $r=\infty$. So the conclusion $\Psi_\pm=0$ again follows from the $L^2$ pairing and integration by parts.

§.§.§ Modes with $l=1$

We now let $k^2=l(l+1)-1=1$. Suppose $(\dg, \dA)$ is the vector perturbations of the following form

\begin{equation}\label{eq:vperturbation_l1}
\dg=\begin{pmatrix}
\end{pmatrix},\quad \dA=\begin{pmatrix}
\end{pmatrix}\quad\mbox{with}\quad \sfV\in\BFV_l ~(l=1).
\end{equation}

We notice that the pure gauge solutions take the form

\begin{equation}
\bdelta\dg=2\delta_g^*\omega=\begin{pmatrix}
0	\\
\end{pmatrix},\quad\!\! \bdelta\dA=\mathcal{L}_{\omega^\sharp}A=0
\end{equation}

\begin{equation}
\omega=\begin{pmatrix}
\end{pmatrix}
\end{equation}

where $L\in C^\infty({\hX})$. When adding $(\bdelta\dg, \bdelta\dA)$ to $(\dg, \dA)$, the quantities $ f,  H_T, K$ change by

\begin{equation}
\bdelta f=r\hd (r^{-1}L), \quad \bdelta K=0.
\end{equation}

Then we define the following gauge-invariant quantities

\begin{equation}
\hd(r^{-1}f)\in C^\infty(\hX; \Lambda^2T^*\hX),\quad K\in C^\infty(\hX)
\end{equation}

If $\hd(r^{-1}f)=0, K=0$, since $\hX$ is contractible, we can find $L\in C^\infty(\hX)$ such that $r^{-1}f=\hd(r^{-1}L)$, which implies

\begin{equation}
(\dg, \dA)=(2\delta^*\omega, \mathcal{L}_{\omega^\sharp}A) \quad\!\! \mbox{with}\quad\!\! \omega=\begin{pmatrix}
\end{pmatrix}
\end{equation}

is a pure gauge solution. As explained around (<ref>) and (<ref>) again, we have $L\in e^{-i\sigma t_*}\eHb^{\infty, \ell}(\CX)$ for $\sigma\neq 0$, while $L\in \eHb^{\infty, \ell-1}(\CX)$ for $\sigma=0$.

Again we can express the linearized Einstein-Maxwell system in terms of the gauge-invariant quantities $\hd(r^{-1}f), K$ as follows

\begin{align}
rf^E&=r^{-2}\hdelta(r^4\hd(r^{-1}f))-4\BQ r^{-2}\hstar\hd(rK)=0,\label{eq:EfVl1}\\
rK^E&=-\hBox(rK)+\frac{2}{r}K+\BQ \hstar\hd(r^{-1}f)=0.\label{eq:EKVl1}
\end{align}

Before solving the above system, we first describe the linearized Kerr-Newman solution in which we linearize the Kerr-Newman solution family $(g_b, A_b)$ around Reissner-Nordström solution $(g_{b_0}, A_{b_0})$ with respect to the angular momentum parameter $\Ba$ in the direction $\dot{\Ba}$, which  is parallel to the rotation axis of $\sfV$. We see that in the static coordinates

\begin{equation}\label{eq:linearizedKNa}
g'_{(\Bm, 0, \BQ)}(0, \dot{\Ba}, 0)=2(\weight-1)\sin^2\theta dtd\varphi,\quad
A'_{(\Bm, 0, \BQ)}(0, \dot{\Ba}, 0)=-\frac{\BQ}{r}\sin^2\theta d\varphi
\end{equation}

where $\abs{\dot{\Ba}}=1$ and $(\theta, \varphi)$ are the spherical coordinates adapted to $\dot{\Ba}$, i.e, $\dot{\Ba}$ is defined by $\theta=0$. We note that (<ref>) is of the form (<ref>) with $f=(-2\Bm r^{-2}+\BQ^2r^{-3})dt, K=-\BQ r^{-2}$ and $\sfV=\sin^2\theta d\phi$. Then the associated gauge-invariant quantities are $\hstar\hd(r^{-1}f)=6\Bm r^{-4}-4\BQ^2r^{-5}, K=-\BQ r^{-2}$.

Now we turn to solving the equations (<ref>) and (<ref>). First (<ref>) implies $\hd\Bigl(r^4\hstar\hd(r^{-1}f)-4\BQ rK\Bigr)=0$ and thus

\begin{equation}\label{eq:solnoff}
r^4\hstar\hd(r^{-1}f)-4\BQ rK=c
\end{equation}

for some constant $c$. Plugging this into the equation (<ref>) yields

\[
(\hBox-V)(rK)=\frac{c\BQ}{r^4}\quad \mbox{where}\quad V=\frac{2}{r^2}+\frac{4\BQ^2}{r^4}.
\]

Since $V>0$ and the right-handed side is stationary, as discussed previously, $rK=0, c=0$ is the only mode solution with frequency $\sigma\neq 0$ to the above equation, and thus $\hd(r^{-1}f)=0$ as well. As for the case $\sigma=0$, since $V=2\rho^2+4\BQ^2\rho^4$ where $\rho=1/r$, using the Frobenius method and analyzing the indicial equation, it follows that for each $c\in\BR$, there is at most one solution $K\in\eHb^{\infty, \ell}(\CX)$ for $-3/2<\ell<-1/2$. On the other hand, one can verify
\[K=-\frac{c\BQ}{6\Bm r^2}\in\eHb^{\infty, \ell}(\CX)
\]
is indeed a solution. Returning to (<ref>) we find
\[\hstar\hd(r^{-1}f)=\frac{c}{6\Bm}(\frac{6\Bm}{r^4}-\frac{4\BQ^2}{ r^5}).
\]

Therefore, replacing $(\dg, \dA)$ by $(\dg, \dA)-\frac{c}{6\Bm}(g', A')$, we may assume $c=0$ and thus $K=0, \hd(r^{-1}f)=0$, which implies $(\dg, \dA)$ is a pure gauge solution.

This finishes the proof of Theorem <ref>.

§ MODE ANALYSIS OF THE SCALAR WAVE OPERATOR

In this section we shall discuss the modes of the scalar wave operator on slowly rotating KN spacetimes with subextremal charge.  It not only occurs as the gauge propagation operator acting on the gauge function $D_{(g_b,A_b)}\widetilde{\Upsilon}_M(\dg,\dA;g_b,A_b)=-\delta_{g_b}\dA$, but an operator on a scalar function which generate the pure gauge solutions to the Maxwell equation. We define the Fourier-transformed scalar wave operator on slowly rotating Kerr-Newman spacetime $g_b$ with parameters $b=(\Bm, \Ba, \BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$ as
\[
\widehat{\Box_{g_b}}(\sigma):=e^{i\sigma t_{b,*}}\Box_{g_b}e^{-i\sigma t_{b,*}}
\]

where $ t_{b,*}=\chi_0(r)(t+r_{b_0,*})+(1-\chi_0(r))(t-r_{(\Bm,0,\BQ),*})$ is defined as in (<ref>).

For ease of notation, we drop $\BC$ in the notations of the functions space $\eHb^{s,\ell}
(\CX;\BC), \eHb^{s,\ell}
(\CX;\BC),\mathcal{A}(\CX;\BC)$ and $C^\infty(\CX;\BC)$ if it is clear from the context.

Let $b_0=(\Bm_0, 0,\BQ_0)$ and $g=g_{b_0}$ be the RN metric with subextremal charge. There exists $\epsilon>0$ such that for $b=(\Bm,\Ba,\BQ)$ with $\abs{b-b_0}<\epsilon$, the following holds

* If $\IM \sigma\geq0$ and $\sigma\neq 0$,

\begin{equation}\label{eq:scalarmodenonzero}
\widehat{\Box_{g_b}}(\sigma):\{u\in\eHb^{s, \ell}(\CX;\BC): \widehat{\Box_g}(\sigma)u\in\eHb^{s,\ell+1}(\CX; \BC)\}\to \eHb^{s,\ell+1}(\CX;\BC)
\end{equation}

is invertible for $s>\frac 12, \ell<-\frac12, s+\ell>-\frac 12$.

* If $s>\frac 12$ and $-\frac 32<\ell<-\frac 12$, then stationary operator

\begin{equation}\label{eq:scalarmodezero}
\widehat{\Box_{g_b}}(0):\{u\in\eHb^{s, \ell}(\CX;\BC): \widehat{\Box_g}(0)u\in\eHb^{s-1,\ell+2}(\CX; \BC)\}\to \eHb^{s-1,\ell+2}(\CX;\BC)
\end{equation}

is invertible.

The proof closely follows <cit.>. We first prove (<ref>) and (<ref>) for the RN metric $g$ and then use a perturbation argument to extend them to the slowly rotating KN metric $g_b$ with $b$ near $b_0$.

* Injectivity of $\widehat{\Box_g}(0)$. Suppose $u\in\eHb^{s,\ell}(\CX)$ with $s>\frac 12,\ell\in(-\frac 32, -\frac 12)$ and

\[
\widehat{\Box_g}(0)u=\Box_gu=\frac{1}{r^2}\pa_r\weight r^2\pa_ru+\frac{1}{r^2}\sL u=0,\quad \weight=1-\frac{2\Bm_0}{r}+\frac{\BQ^2_0}{r^2}.	\]

Then by Proposition <ref>, $u\in\rho C^\infty(\CX)+\mathcal{A}^{2-}(\CX)$ where $\rho=r^{-1}$, which implies $\abs{u},\abs{r\pa_r u},\abs{\snabla u}\lesssim r^{-1}$. As a consequence, the boundary term at $r=\infty$ in the following integration by parts

\begin{equation}\label{eq:ibpzero}
0=-\int_{\BS^2}\!\int_{\ehRN}^\infty\!\Box_gu\cdot\overline{u}\,r^2drd\sg=\int_{\BS^2}\!\int_{\ehRN}^\infty\!\Bigl(\weight(r)\abs{r\pa_r u}^2+\abs{\snabla u}^2\Bigr)\,drd\sg
\end{equation}

vanishes, and the boundary term at $r=\ehRN$ vanishes since $u$ is smooth there and $\weight(\ehRN)=0$. Then $u$ is a constant in $r\geq \ehRN$ and thus vanishes there since $\lim_{r\to\infty}u=0$. Since $u$ vanishes to infinite order at $r=\ehRN$ and smooth in $r\leq \ehRN$, we shall prove that $u=0$ in $r\leq \ehRN$ as well by following the arguments in <cit.>. Concretely, we first compute the following energy identity in $r_-\leq r<\ehRN$

\begin{equation}\label{eq:energyidentity}
\begin{split}
&\qquad\qquad\pa_r\Bigl(\abs{r-\ehRN}^{-N}\bigl(-\weight(r)\abs{\pa_r u}^2+\frac{1}{r^2}\abs{\sd u}_{\sg}^2+\abs{u}^2\bigr)\Bigr)+2\abs{r-\ehRN}^{-N}\frac{1}{r^2}\sdelta\big(\RE(\overline{\pa_r u}\sd u)\big)\\
&\qquad\qquad=N\abs{r-\ehRN}^{-N-1}\bigl(-\weight(r)\abs{\pa_r u}^2+\frac{1}{r^2}\abs{\sd u}_{\sg}^2+\abs{u}^2\bigr)-\abs{r-\ehRN}^{-N}\bigl(2\RE(\overline{\pa_r u}\cdot\widehat{\Box_g}(0)u)-R(u, du)\bigr)
\end{split}
\end{equation}

\[
R(u, du)=\bigl(4r^{-1}\weight(r)-\weight'(r)\bigr)\abs{\pa_r u}^2+2\RE(\overline{u}\pa_r u)-2r^{-3}\abs{\sd u}_{\sg}^2
\]

is a quadratic form in $u$ and $du$ independent of $N$. As a result, $R(u, du)$ can be controlled by the first term on the right-hand side of the above energy identity (<ref>) when $N$ is sufficiently large. Then for any fixed $r_0\in[r_-, \ehRN)$ we apply Stokes' Theorem in $[r_0, \ehRN-\epsilon]\times\BS^2$. For $N$ large enough we have

\begin{equation}\label{eq:Stokesthm}
\begin{split}
&\int_{\BS^2}\!(-\weight(r)\abs{\pa_r u}^2+\abs{\sd u}^2+\abs{u}^2)|_{r=r_0}\,d\sg\\
&\quad\leq (\ehRN-r_0)^N\epsilon^{-N}\int_{\BS^2}\!(-\weight(r)\abs{\pa_r u}^2+\abs{\sd u}^2+\abs{u}^2)|_{r=\ehRN-\epsilon}\,d\sg\leq C_K\epsilon^{-N+K}
\end{split}
\end{equation}

for any $K$ as $\epsilon\to 0^+$ since $u$ is smooth and vanishes to infinite order at $r=\ehRN$. By choosing $K>N$ and letting $\epsilon\to0^+$ we conclude that $u=0$ at $r=r_0$.

* Surjectivity of $\widehat{\Box_g}(0)$. The index $0$ property of $\widehat{\Box_{g}}(0)$ established in Lemma <ref> directly gives the surjectivity. But here we present an alternative proof of the surjectivity. To prove the surjectivity of $\widehat{\Box_g}(0)$, we note that it is equivalent to proving the injectivity of the adjoint $\widehat{\Box_g}(0)^*$. Suppose $v\in \sHb^{-s+1, -\ell-2}(\CX)$ with $s>\frac 12,\ell\in(-\frac 32, -\frac 12)$ satisfies $\widehat{\Box_g}(0)^* v=\widehat{\Box_g}(0)v=0$. By Proposition <ref>, $v$ is smooth in $r_-\leq r<\ehRN$ and $r>\ehRN$, and $u\in\rho C^\infty(\CX)+\mathcal{A}^{2-}(\CX)$ near $r=\infty$. First we have $v=0$ in $r<\ehRN$ since $\widehat{\Box_g}(0)^*$ is a hyperbolic operator there with $r$ being the timelike function, see <cit.>. Next, according to <cit.>, $v$ is conormal at $r=\ehRN$. Since the normal operator of $\weight\widehat{\Box_g}(0)^*$ is $(\weight\pa_r)^2$ whose boundary spectrum consists of $\{(0,0), (0,1)\}$,
Once we know that $v$ is conormal at $r=\ehRN$, say $u\in\mathcal{A}^l([\ehRN,\infty)\times\BS^2)$ for some $l\in\BR$, which can be identified with $C^\infty(\BS^2;\mathcal{A}^l([\ehRN,\infty))$. Then we can treat $\omega\in\BS^2$ as a parameter and solve the PDE $\widehat{\Box_g}(0)^* v=0$ as an ODE with respect to $r$ as follows

\begin{equation}
(\weight r^2\pa_r)u=-\weight r^2\sL u\in \mathcal{A}^{l+1}.
\end{equation}

This ODE has regular singular point at $r=\ehRN$ with indicial root being $0$ with multiplicity $2$. If $l+1<0$, we conclude that $v\in\mathcal{A}^{l+1}$ which improves the decay as $r\to \ehRN^+$ by $(r-\ehRN)$.

it follows that $v=H(r-\ehRN)(v_0+v_1\log\weight+\tilde{v})$ where $v_0, v_1\in C^\infty(\BS^2)$ and $\tilde{v}\in\mathcal{A}^{1-}([\ehRN,\infty))$ which means that  $\tilde{v}$ is conormal at $r=\ehRN$ and $\tilde{v}\sim (r-\ehRN)^{1-}$ as $r\to\ehRN^+$. Since $\widehat{\Box_g}(0)H(r-\ehRN)=r^{-2}\pa_r(r^2\weight(r)\delta(r-\ehRN))=0$ and $\widehat{\Box_g}(0)(H(r-\ehRN)\log\weight)=\weight'(r)\delta(r-\ehRN)+2\BQ_0^2r^{-4}H(r-\ehRN)$, we have

\begin{align*}
\end{align*}

\[
R=\frac{H(r-\ehRN)}{r^2}\bigl(\sL v_0+\log(r-\ehRN)\sL v_1+\frac{2\BQ_0^2}{r^2}v_1\bigr)+H(r-\ehRN)\widehat{\Box_g}(0)\tilde{v}\in\mathcal{A}^{0-}.
\]

However, $\delta(r-\ehRN)\notin \mathcal{A}^{0-}$, and this implies $v_1=0$. Since $v_1=0$ implies $\weight\pa_r v\to 0$ as $r\to \ehRN^+$, we find that the integration by parts (<ref>) still works here and thus gives $v=0$ in $r>\ehRN$. Then we can conclude that $v$ is supported at $r=\ehRN$, and consequently $v\notin L^2$ <cit.> unless $v=0$. However, the radial point estimates at the event horizon imply that $v\in H^{1/2-}$ near $r=\ehRN$. Therefore, $v$ must be $0$.

* Invertibility of $\widehat{\Box_g}(\sigma)$ for $\sigma\in\BR\setminus\{0\}$. According to Lemma <ref>, it suffices to prove the injectivity of $\widehat{\Box_g}(\sigma)$ for non-zero real $\sigma$. We first illustrate the relationship between $\widehat{\Box_g}(\sigma)$ and $\wt{\Box_g}(\sigma)=e^{i\sigma t}\Box_g e^{-i\sigma t}$, that is, $u\in\ker\widehat{\Box_g}(\sigma)$ gives rise to an outgoing solution

\begin{equation}\label{eq:differentconjBox}
\wt{\Box_g}\tilde{u}=0\quad \mbox{where}\quad  \tilde{u}=e^{i\sigma(t-t_{b_0,*})}u.
\end{equation}

We next prove that $\tilde{u}=0$ and thus $u=0$ in $r\geq \ehRN$ by a boundary pairing argument (see<cit.><cit.><cit.>), and a unique continuation theorem at infinity. Concretely, let $f(r)\in C^\infty([\ehRN, \infty))$ be a nonnegative function with $f(r)=r-\ehRN$ for $\ehRN\leq r<3\Bm_0$ and $f(r)=r^{-1}$ for $r>4\Bm_0$. Fix a cutoff $\chi\in C^\infty([0,\infty))$ which is $0$ on $[0,1/2]$ and $1$ on $[1,\infty)$. For small $\epsilon>0$, we define

\begin{equation}\label{eq:cutoff}
\chi_\epsilon(r):=\chi(\frac{f}{\epsilon})\in C^\infty_c((\ehRN, \infty)).
\end{equation}

We see that $\chi_\epsilon=1$ when $r-\ehRN\geq \epsilon, r\leq\epsilon^{-1}$, and $r-\ehRN\geq \epsilon/2, r\leq 2\epsilon^{-1}$ on supp$\chi_\epsilon$. We then calculate

\begin{equation}\label{eq:boundarypair}
\begin{split}
&=\lim_{\epsilon\to 0}\int_{\BS^2}\!\int_{\ehRN}^\infty\![\wt{\Box_g}(\sigma),\chi_\epsilon]\tilde{u}\cdot\overline{\tilde{u}}\,r^2drd\sg\\
&=\lim_{\epsilon\to 0}\Bigl(\int_{\BS^2}\!\int_{\ehRN}^\infty\!2\weight\pa_r\chi_{\epsilon}\cdot\pa_r \tilde{u}\cdot\overline{\tilde{u}}\,r^2drd\sg-\int_{\BS^2}\!\int_{\ehRN}^\infty\!\weight\pa_r\chi_{\epsilon}\cdot\pa_r \abs{\tilde{u}}^2\,r^2drd\sg\Bigr).
\end{split}
\end{equation}

where $\pa_r\chi_\epsilon$ may be nonzero only near the event horizon $r=\ehRN$ and near infinity $r=\infty$. Near $r=\ehRN$, we have $\tilde{u}=e^{-i\sigma r_*}u$ where we write $r_*=r_{b_0, *}$ for brevity of notation,  $\pa_r\chi_{\epsilon}=\epsilon^{-1}\chi'(\frac{r-\ehRN}{\epsilon})\to\delta(r-\ehRN)$ as $\epsilon\to 0$, and thus

\begin{equation}\label{eq:commutatoratEH}
\begin{split}
&\lim_{\epsilon\to 0}\Bigl(\int_{\BS^2}\!\int_{\ehRN}^\infty\!2\weight\pa_r\chi_{\epsilon}\cdot\pa_r \tilde{u}\cdot\overline{\tilde{u}}\,r^2drd\sg-\int_{\BS^2}\!\int_{\ehRN}^\infty\!\weight\pa_r\chi_{\epsilon}\cdot\pa_r \abs{\tilde{u}}^2\,r^2drd\sg\Bigr)\\
&\quad=\int_{\BS^2}\!\Big(-2i\sigma r^2\abs{u}^2+2\weight r^2\pa_r u\cdot \overline{u}-r^2\weight\pa_r\abs{u}^2\Big)|_{r=\ehRN}\,d\sg\\
&\quad=\int_{\BS^2}\!\big(-2i\sigma r^2\abs{u}^2\big)|_{r=\ehRN}\,d\sg\\
\end{split}
\end{equation}

where we use the smoothness of $u$ at $r=\ehRN$ in the last step. While near $r=\infty$, we have $r^2\pa_r\chi_{\epsilon}=-\epsilon^{-1}\chi'(\frac{\rho}{\epsilon})\to-\delta(\rho)$ with $\rho=r^{-1}$ as $\epsilon\to 0$, $\tilde{u}=e^{i\sigma r_*}u$ where $u=\rho u_+(\omega)+e(\rho,\omega)$ with $u_+(\omega)\in C^\infty(\BS^2)$ and $e(\rho, \omega)\in\mathcal{A}^{2-}(\CX)$ (see Proposition <ref>), and thus

\begin{equation}\label{eq:commutatoratinfty}
\begin{split}
&\lim_{\epsilon\to 0}\Bigl(\int_{\BS^2}\!\int_{\ehRN}^\infty\!2\weight\pa_r\chi_{\epsilon}\cdot\pa_r \tilde{u}\cdot\overline{\tilde{u}}\,r^2drd\sg-\int_{\BS^2}\!\int_{\ehRN}^\infty\!\weight\pa_r\chi_{\epsilon}\cdot\pa_r \abs{\tilde{u}}^2\,r^2drd\sg\Bigr)\\
&\quad=\int_{\BS^2}\!\Big(-2i\sigma r^2\abs{u}^2-2\weight r^2\pa_r u\cdot \overline{u}+r^2\weight\pa_r\abs{u}^2\Big)|_{r=\infty}\,d\sg\\
&\quad=\int_{\BS^2}\!\big(-2i\sigma r^2\abs{u}^2\big)|_{r=\infty}\,d\sg=\int_{\BS^2}\!-2i\sigma \abs{u_+}^2\,d\sg\\
\end{split}
\end{equation}

Putting (<ref>) and (<ref>) together yields

\begin{equation}
0=-2i\sigma\Bigl(\int_{\BS^2}\! r^2\abs{u}^2|_{r=\ehRN}\,d\sg+\int_{\BS^2}\!\abs{u_+}^2\,d\sg\Bigr),
\end{equation}

and this implies $u|_{r=\ehRN}=0$ and $u_+(\omega)=0$. This proves that the leading order term of $u$ at infinity vanishes and thus $u\in\mathcal{A}^{2-}(\CX)$. For the next step, we rewrite $\widehat{\Box_g}(\sigma)u=0$ near infinity as follows

\[
2i\sigma\pa_r (ru)=-\frac{1}{r}\pa_rr^2\weight\pa_ru-\frac{1}{r}\sL u=\rho\mbox{Diff}_b^2u.
\]

Then integrating both sides from infinity to $r$ and using $ru|_{r=\infty}=0$ yield $u\in\mathcal{A}^{3-}(\CX)$. Proceeding iteratively, we find that $u\in\mathcal{A}^{\infty}$, which means that $u$ vanishes to infinite order at $r=\infty$. Then using the unique continuation theorem at infinity <cit.> for $\wt{\Box_g}(\sigma)\tilde{u}=0$ gives $\tilde{u}=0$ and thus $u=0$ in $r>\ehRN$. By Proposition <ref> again, $u$ is smooth and vanishes to infinite order at $r=\ehRN$. Therefore, we follow the arguments in <cit.> again to prove $u=0$ in $r<\ehRN$. Since we are working in the region near the $r=\ehRN$ where $t_{b_0,*}=t_0$, it would be more convenient to use the coordinates $(t_{b_0}, r,\omega)$ with $\omega\in\BS^2$, and then we write

\begin{equation}
\widehat{\Box_g}(\sigma)u=\frac{1}{r^2}\pa_rr^2\weight\pa_ru+\frac{1}{r^2}\sL u-\frac{2i\sigma}{r}\pa_rr u.
\end{equation}

We can again obtain the energy identity as in (<ref>) with $R(u, du)$ replaced by
\[R(u,du)-2\RE\bigl(2i\sigma r^{-1}\overline{\pa_ru}\cdot\pa_r(ru)\bigr)
\]
which can still be controlled when $N$ is sufficiently large. Proceeding in the same way as in (<ref>), it follows that $u=0$ in $r<\ehRN$ as well.

* Invertibility of $\widehat{\Box_g}(\sigma)$ for $\IM\sigma>0$. Again according to Lemma <ref>, it suffices to prove the injectivity of $\widehat{\Box_g}(\sigma)$ for $\sigma\!\in\!\BC, \IM\sigma\!\!>0$. We again use relationship between $\widehat{\Box_g}(\sigma)$ and $\wt{\Box_g}(\sigma)$ as illustrated in (<ref>). Now $\tilde{u}$ decays exponentially when $r_{b_0, *}\to\pm\infty$. So we can apply integration by parts to conclude

\[
0=-\int_{\BS^2}\!\int_{\ehRN}^\infty\!\wt{\Box_g}(\sigma)\tilde{u}\cdot\overline{\tilde{u}}\,r^2dr\sg=\int_{\BS^2}\!\int_{\ehRN}^\infty\!\Bigl(\weight(r)\abs{r\pa_r \tilde{u}}^2+\abs{\snabla \tilde{u}}^2-\sigma^2\weight^{-1}(r)\abs{\tilde{u}}^2\Bigr)\,drd\sg.
\]

Then by taking imaginary part for $\RE\sigma\neq 0$, while using $-\sigma^2>0$ for $\RE\sigma=0$, it follows that $\tilde{u}=0$ and thus $u=0$ in $r>\ehRN$. The proof of $u=0$ in $r<\ehRN$ is the same as that in the non-zero real $\sigma$ case.

* $\widehat{\Box_{g_b}}(\sigma)$ has index $0$ for $\IM\sigma\geq 0, \sigma\neq 0$. We now prove that $\widehat{\Box_{g_b}}(\sigma)$ with $\IM \sigma\geq 0, \sigma\neq 0$ has index $0$, from which the surjectivity of $\widehat{\Box_g}(\sigma)$ with $\IM \sigma\geq 0, \sigma\neq 0$ follows. In view of the high energy estimates in Theorem, $\widehat{\Box_{g_b}}(\sigma)$ is invertible and thus has index $0$ for fixed $\IM\sigma=C\geq 0$ when $\abs{\RE\sigma}\gg1$. We claim that the index of $\widehat{\Box_{g_b}}(\sigma)$ with $\IM \sigma\geq 0,\sigma\neq 0$ is a constant on the horizontal line $\IM\sigma=C$ and thus is $0$ for all $\sigma$ with $\IM\sigma\geq 0,\sigma\neq0$. Let

\[
\mathcal{X}(\sigma):=
\{u\in\eHb^{s,\ell}(\CX):\widehat{\Box_{g_b}}(\sigma)u\in \eHb^{s,\ell+1}(\CX)\}.
\]

For any fixed $\sigma_0\neq 0$ with $\IM\sigma_0=C\geq0$, without loss of generality, we may assume $\RE \sigma_0\geq 0$. First, there exists a $C'$ such that $\widehat{\Box_{g_b}}(\sigma)$ is invertible and thus has index $0$ for $\sigma\in[C'+iC, \infty+iC)$. We next prove that for each $\sigma'\in[\sigma_0=\RE\sigma_0+iC, C'+iC]$, there exists $\epsilon>0$ such that the index of $\widehat{\Box_{g_b}}(\sigma)$ is a constant for $\sigma$ with $\IM\sigma\geq 0,\abs{\sigma-\sigma'}<\epsilon$. Then by compactness of $[\sigma_0=\RE\sigma_0+iC, C'+iC]$ we conclude that the index of $\widehat{\Box_{g_b}}(\sigma_0)$ is the same as that of $\widehat{\Box_{g_b}}(C'+iC)$. Suppose $\widehat{\Box_{g_b}}(\sigma')$ has kernel and cokernel of dimension $k_+$ and $k_-$, respectively, then we establish a Grushin problem <cit.>. More specifically, we define the following operator

\[
\widehat{\Box_{g_b}}(\sigma)&L_-\\
\end{pmatrix}: \mathcal{X}(\sigma)\oplus\BC^{k_-}\to \eHb^{s,\ell+1}(\CX)\oplus \BC^{k_+}
\]

where $L_+, L_-$ are defined as follows. If $\ker\widehat{\Box_{g_b}}(\sigma')$ is spanned by $x_j\in \mathcal{X}(\sigma'), j=1,\cdots, k_+$, then by Hahn-Banach Theorem  we can find $x_j^*\in \sHb^{-s,-\ell}(\CX)$ with $\norm{x_j^*}\leq 1$ such that $\angles{x_i}{x_j^*}=\delta_{ij}$. It follows that

\[
L_+: \eHb^{s,\ell}(\CX)\to C^{k_+},\quad L_+(f):=(\angles{f}{x_1^*},\cdots,\angles{f}{x_{k_+}^*})
\]

restricts to an isomorphism $\ker\widehat{\Box_{g_b}}(\sigma')\to \BC^{k_+}$. As for the construction of $L_-$, we choose $y_j\in \eHb^{s,\ell+1}(\CX)$ with $\norm{y_j}\leq 1$, $ j=1, \cdots, k_-$ so that $y_j+\Ran\widehat{\Box_{g_b}}(\sigma')$ form a basis of $\eHb^{s,\ell+1}/\Ran\widehat{\Box_{g_b}}(\sigma')$. Then we define

\[
L_-:\BC^{k_-}\to \eHb^{s,\ell+1}(\CX),\quad L_-((a_1, \cdots,a_{k_-})):=\sum_{j=1}^{k_-}a_jy_j.
\]

The operator $L_-$ has maximal rank and $\Ran L_-\cap \Ran\widehat{\Box_{g_b}}(\sigma')=\emptyset$. The above construction of $L_+,L_-$ implies $L(\sigma')$ has a trivial kernel and onto, and thus is invertible. Since $\widehat{\Box_{g_b}}(\sigma)$ satisfy the uniform Fredholm estimates for all $\sigma\in\BC, \IM\sigma\in[0,2C]$ satisfying $\abs{\sigma_0}/2\leq \abs{\sigma}\leq 2(C+C') $, so do $L(\sigma)$. That is, there exists a constant $\tilde{C}$ (independent of $b$ and $\sigma$) such that

\[
\norm{(u, u^-)}_{\eHb^{s,\ell}(\CX)\oplus \BC^{k_-}}\leq \tilde{C}\Bigl(\norm{L(\sigma)(u, u^-)}_{\eHb^{s,\ell+1}(\CX)\oplus \BC^{k_+}}+\norm{(u,u^-)}_{\eHb^{s_0,\ell_0}(\CX)\oplus \BC^{k_-}}\Bigr)
\]

for $\sigma\in\BC, \IM\sigma\in[0,2C]$ satisfying $\abs{\sigma_0}/2\leq \abs{\sigma}\leq 2(C+C')$, and $s_0<s, \ell_0<\ell$. Then following the perturbation arguments in <cit.>, we prove the invertibility of $L(\sigma)$ when $\sigma$ is close to $\sigma'$. Concretely, suppose there exists a sequence $\delta_j\to\delta'$ such that $L(\delta_j)$ is not invertible, so either $\ker L(\delta_j) $ on $\eHb^{s,\ell}(\CX)\oplus \BC^{k_-}$ or $\ker L(\delta_j)^* $ on $\sHb^{-s,-\ell-1}(\CX)\oplus \BC^{k_+}$ is non-trivial. By passing to a subsequence, we may assume that the former possibility holds for all $j$, as the case of the adjoint is completely analogous. Now, if $\norm{(u_j, u^-_j)}_{\eHb^{s,\ell}(\CX)\oplus \BC^{k_-}}=1$ and $L(\delta_j)(u_j, u^-_j)=0$, then the above uniform Fredholm estimate gives $\tilde{C}^{-1}\leq \norm{(u,u^-)}_{\eHb^{s_0,\ell_0}(\CX)\oplus \BC^{k_-}}$. Using the sequential compactness of the unit ball in $\eHb^{s,\ell}(\CX)\oplus \BC^{k_-}$ in the weak topology, and the compactness of the inclusion $\eHb^{s,\ell}(\CX)\oplus \BC^{k_-}\to\eHb^{s_0,\ell_0}(\CX)\oplus \BC^{k_-}, s_0<s,\ell_0<\ell$,  it follows that $(u_j, u^-_j)$ has a weakly convergent subsequence in $\eHb^{s,\ell}(\CX)\oplus \BC^{k_-}$ to some $(u_0, u^+_0) \in \eHb^{s,\ell}(\CX)\oplus \BC^{k_-}$ which is  norm-convergent in $\eHb^{s_0,\ell_0}(\CX)\oplus \BC^{k_-}$. We also have

\[
\qquad \qquad L(\delta')(u_0, u^+_0)=L(\delta')\left((u_0, u^+_0)-(u_j, u^-_j)\right)+(L(\delta')-L(\delta_j))(u_j, u^-_j)\to 0\quad\mbox{in}\quad \eHb^{s_0-2, \ell_0}(\CX)\oplus\BC^{k_-}
\]

since $L(\delta_j)\to L(\delta')$ as bounded operators in $\mathcal{L}(\eHb^{s_0, \ell_0}(\CX)\oplus\BC^{k_-}, \eHb^{s_0-2, \ell_0}(\CX)\oplus\BC^{k_-})$ and $(u_j, u^-_j)\to(u_0, u^-_0)$ in $\eHb^{s_0, \ell_0}(\CX)\oplus\BC^{k_-}$. Therefore, $L(\delta')(u_0, u^-_0)=0$. Since $\tilde{C}^{-1}\leq \norm{(u,u^-)}_{\eHb^{s_0,\ell_0}(\CX)\oplus \BC^{k_-}}$, $(u_0, u^-_0)\neq 0$ and thus $\ker L(\sigma')$ is non-trivial, which leads to a contradiction. This proves that $L(\sigma)$ is invertible for $\sigma$ close to $\sigma'$, which implies $\widehat{\Box_{g_b}}(\sigma)$ has index $k_+-k_-$ <cit.>.

* Proof for $\widehat{\Box_{g_b}}(\sigma)$. Finally, we consider the slowly rotating KN metric $g_b$, with $b$ near $b_0$. By Theorem <ref> and high energy estimate <ref>, $\widehat{\Box_{g_b}}(\sigma)$ is invertible for $\abs{\sigma}$ large, say $\abs{\sigma}\geq C$, when $\abs{b-b_0}<\epsilon_1$, so we only need to consider the bounded region$\abs{\sigma}\leq C$. Near $\sigma=0$, the invertibility of $\widehat{\Box_{g_b}}(\sigma)$ for $b$ close to $b_0$ follows from the perturbation argument as above and the fact that $\widehat{\Box_{g}}(0)$ is invertible. For $\sigma\in\BC, \IM\sigma\geq 0$, bounded away from $0$ and infinity, we prove the theorem by perturbation arguments (starting with the invertibility of $\widehat{\Box_{g}}(\sigma) $) in a compact set of $\IM\sigma\geq0$. Concretely, suppose there exist sequences $\delta_j\to 0, b_j\to b_0$ such that $\ker\widehat{\Box_{g_{b_j}}}(\sigma_j)$ is non-trivial. Then we can find $u_j\in\eHb^{s,\ell}(\CX)\cap\ker\widehat{\Box_{g_{b_j}}}(\sigma_j)$ with $\norm{u_j}_{\eHb^{s,\ell}(\CX)}=1$. By the uniform Fredholm estimate <ref>, we have $1\leq C'\norm{u_j}_{\eHb^{s_0, l_0}(\CX)}$. Therefore, there exists a subsequence $u_j\to u$ weakly in $\eHb^{s, \ell}(\CX)$ which is norm convergent in $\eHb^{s_0, \ell_0}(\CX)$ with the limit $u$ being non-zero and satisfying $\widehat{\Box_{g}}(0)u=0$. But this contradicts the invertibility of $\widehat{\Box_{g}}(0)$. So this proves the injectivity. The surjectivity which is equivalent to the injectivity of the adjoint $\widehat{\Box_{g_{b}}}(\sigma)^*$ can be proved in a similar manner. This proves that for $s>\frac 12,\ell\in (-\frac 32, \frac 12)$ and $\IM\sigma\geq 0,\abs{\sigma}<c, \abs{b-b_0}<\epsilon_2$ for some $c, \epsilon_2>0$,

\[\widehat{\Box_{g_{b}}}(\sigma):\{u\in\eHb^{s,\ell}(\CX):\widehat{\Box_{g_{b}}}(\sigma)u\in\eHb^{s-1,\ell+2}(\CX)\}\to \eHb^{s-1,
\ell+2}(\CX)
\]

is invertible. By Proposition <ref>, this implies that $\widehat{\Box_{g_{b}}}(\sigma)$ acting on the function spaces in (<ref>) is injective for $\IM\sigma\geq0, 0<\abs{\sigma}<c$, and the invertibility follows from the fact that $\widehat{\Box_{g_{b}}}(\sigma)$ has index $0$. This proves the theorem for $\IM\sigma\geq 0,0\leq\abs{\sigma}<c$ and $\abs{b-b_0}<\epsilon_2$. For $\IM\sigma\geq 0, c\leq\abs{\sigma}\leq C$, the fact that $\widehat{\Box_g}(\sigma)$ is invertible and a perturbation argument as above imply the invertibility of $\widehat{\Box_{g_b}}(\sigma)$ for $(b, \sigma)$ in an open neighborhood of $(b_0, \sigma)$. Then the compactness of the region $c\leq\abs{\sigma}\leq C$ implies that there exists an $\epsilon_3>0$ (depending on $c, C$) such that $\widehat{\Box_{g_b}}(\sigma)$ in (<ref>) is invertible for $\IM\sigma\geq 0, c\leq\abs{\sigma}\leq C$ and $\abs{b-b_0}<\epsilon_3$. Therefore, the theorem holds for $\abs{b-b_0}<\epsilon$ where $\epsilon=\min\{\epsilon_1, \epsilon_2, \epsilon_3\}$.

§.§ Growing zero modes

For later use, we now discuss the explicit form of the scalar functions in $\ker\widehat{\Box_{g_{b}}}(0)$ which are allowed to have more growth at infinity.

For $b=(\Bm, \Ba, \BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$, we have

\begin{align}\label{eq:gscalar3/2}
\ker\widehat{\Box_{g_b}}(0)\cap\eHb^{\infty, -3/2-}(\CX)&=\langle u_{b, s_0}\rangle, \quad u_{b,s_0}=1.\\\label{eq:gscalardual3/2}
\ker\widehat{\Box_{g_b}}(0)^*\cap\sHb^{-\infty, -3/2-}(\CX)&=\langle u^*_{b, s_0}\rangle,\quad u^*_{b,s_0}=H(r-\ehKN).
\end{align}

The following spaces

\begin{align}\label{eq:gscalar5/2}
\ker\widehat{\Box_{g_b}}(0)\cap\eHb^{\infty, -5/2-}(\CX)&=\langle u_{b, s_0}\rangle\oplus\{u_{b, s_1}(\sfS): \sfS\in\BFS_1\},\\\label{eq:gscalardual5/2}
\ker\widehat{\Box_{g_b}}(0)^*\cap\sHb^{-\infty, -5/2-}(\CX)&=\langle u^*_{b, s_0}\rangle\oplus\{u^*_{b, s_1}(\sfS): \sfS\in\BFS_1\},
\end{align}

are $4$-dimensional. Moreover, the maps $b\mapsto u_{b, s_1}(\sfS)$ and $b\mapsto u^*_{b,s_1}(\sfS)$ can be chosen to be continuous in $b$ with values in the respective spaces. More explicitly, we have

\begin{equation}\label{eq:gscalar5/2RN}
u_{b=(\Bm, 0,\BQ),s_1}(\sfS)=(r-\Bm)\sfS,\quad u_{b=(\Bm, 0,\BQ),s_1}(\sfS)=(r-\Bm)H(r-\ehKN)\sfS.
\end{equation}

We use the notation $u_{b, s_0}, u_{b,s_1}$ because they are of scalar type $l=0$ and $l=1$ respectively when $b=(\Bm, 0,\BQ)$ is a parameter of RN metric.

We first prove (<ref>). Let $u\in\ker\widehat{\Box_{g_b}}(0)\cap \eHb^{\infty,-3/2-}(\CX)$. By the normal operator argument as in  Proposition <ref>,
$u$ must have the form $u=u_0+\tilde{u}$ where $u_0\in\BC$ and $\tilde{u}\in\mathcal{A}^{1-}(\CX)\subset\eHb^{\infty, -1/2-}(\CX)$. Then we have

\[
\]

In view of Theorem <ref>, $\tilde{u}$ must be $0$, and thus $u=u_0$ is a constant. Conversely, we certainly have $\widehat{\Box_{g_b}}(0)1=0$. This proves (<ref>).

The proof of (<ref>) is analogous. Let $u\in\ker\widehat{\Box_{g_b}}(0)\cap \sHb^{-\infty,-3/2-}(\CX)$. By the normal operator argument again, we conclude that $u=\chi u_0+\tilde{u}$ where $\chi$ is a smooth cutoff with $\chi=0$ for $r\leq 3\Bm_0$, $\chi=1$ with $r\geq 4\Bm_0$, and $u_0\in\BC, \tilde{u}\in\sHb^{\infty, -1/2-}(\CX)$. We calculate

\[
-\widehat{\Box_{g_b}}(0)^*\chi u_0\in\sHb^{-\infty,\infty}(\CX)\subset\sHb^{-\infty, 3/2-}(\CX).
\]

Owing to Theorem <ref>, there exists a unique $\tilde{u}\in\sHb^{-\infty, -1/2-}(\CX)$ satisfying $\widehat{\Box_{g_b}}(0)^*\tilde{u}=-\widehat{\Box_{g_b}}(0)^*\chi u_0$. This proves that the space in (<ref>) is at most $1$-dimensional. Conversely, a direct calculation implies
\[
\widehat{\Box_{g_b}}(0)H(r-\ehKN)=\rho_b^{-2}\pa_r(\Delta_b\delta(r-\ehKN))=0
\]
since $\Delta_b|_{r=\ehKN}=0$. This proves (<ref>).

As for (<ref>), since $\eHb^{\infty, -5/2-}(\CX)\subset\mathcal{A}^{-1-}(\CX)$, in the normal operator argument as in Proposition <ref>, we shift the contour of integration through the pole $i$ with the space of resonant states given by $r\sfS$ with $\sfS\in\BFS_1$. That is to say, $u$ must take the form $u=\chi r\sfS+\tilde{u}$ where $\chi$ is the smooth cutoff as defined above and $\sfS\in\BFS_1, \tilde{u}\in\eHb^{\infty, -3/2-}(\CX)$. Since $\BFS_1$ is $3$-dimensional, we find that the space in (<ref>) is at most $4$-dimensional. Now we prove that it is indeed $4$-dimensional. We compute

\begin{equation}\label{eq:eqnforerror}
-\widehat{\Box_{g_b}}(0)\chi r\sfS=-\chi\widehat{\Box_{\underline{g}}}(0)r\sfS+[\chi, \widehat{\Box_{\underline{g}}}(0)]r\sfS+(\widehat{\Box_{\underline{g}}}(0)-\widehat{\Box_{g_b}}(0))\chi r\sfS=0+\eHb^{\infty, \infty}(\CX)+\eHb^{\infty, 1/2-}(\CX)
\end{equation}

where we use $\widehat{\Box_{\underline{g}}}(0)-\widehat{\Box_{g_b}}(0)\in\rho^3\mbox{Diff}_b^2$ for the third summand. Since $\ker\widehat{\Box_{g_b}}(0)\cap \sHb^{-\infty, -1/2-}(\CX)=0$, there exists $\tilde{u}\in\eHb^{\infty, -3/2-}(\CX)$, which is unique modulo a multiple of $u_{b, s_0}$, satisfying the above equation (<ref>). Concretely, let $v\in C^\infty_c(X^\circ)$ such that $\langle u_{b,s_0},v\rangle=1$. Then there exists a unique $\tilde{u}\in \ann(v)$ satisfying the equation (<ref>). Therefore, $\chi r\sfS+\tilde{u}$, where $\tilde{u}$ is uniquely determined by $\sfS$ and $b$, indeed gives rise to an element in the space (<ref>) with leading order term $r\sfS$ at infinity. This proves (<ref>). The proof of $\eqref{eq:gscalardual5/2}$ is completely analogous.

Now we turn to discussing the continuous dependence of $u_{b, s_1}(\sfS), u^*_{b, s_1}(\sfS)$ on $b$.
Suppose $b_j\to b$ and $u_{b_j}(\sfS)=\chi r\sfS+\tilde{u}_{b_j}(\sfS)\in\ker\widehat{\Box_{g_{b_j}}}\cap \eHb^{\infty, -5/2-}(\CX)$ where $\tilde{u}_{b_j}(\sfS)$ is in the annihilator of $v\in C^\infty_c(X^\circ)$ which satisfies $\langle u_{b,s_0}, v\rangle \neq 0$. Let $u=\chi r\sfS+\tilde{u}(\sfS)\in\ker\widehat{\Box_{g_b}}(0)$ and $e_j(\sfS)=u_{b_j}(\sfS)-u(\sfS)=\tilde{u}_{b_j}(\sfS)-\tilde{u}(\sfS)$, and we find

\[
\widehat{\Box_{g_b}}(0)e_j(\sfS)=(\widehat{\Box_{g_b}}(0)-\widehat{\Box_{g_{b_j}}}(0))(\tilde{u}_{b_j}(\sfS)+\chi r\sfS).
\]

Since $\{\tilde{u}_{b_j}(\sfS)\}$ is uniformly bounded in $\ann(v)$ and $\widehat{\Box_{g_b}}(0)-\widehat{\Box_{g_{b_j}}}(0)\in (b-b_j)\rho^3\mbox{Diff}_b^2$, we see that $\widehat{\Box_{g_b}}(0)e_j(\sfS)\in (b-b_j)\eHb^{\infty, 1/2-}(\CX)$. As $\widehat{\Box_{g_b}}(0)|_{\ann(v)}: \ann{v}\to\eHb^{\infty,1/2-}(\CX)$ is invertible, it follows that $e_j(\sfS)$ is of size $\mathcal{O}_{\eHb^{\infty, -3/2-}(\CX)}(b-b_j)$. This completes the proof of the continuity.

It remains to derive the explicit expressions in (<ref>). We first consider the case $b=b_0$, i.e. $g_b=g$. Using $\sL\sfS=-2\sfS$, it follow that

\begin{equation}
\begin{split}
\widehat{\Box_{g}}(0)((r-\Bm_0)\sfS)&=\frac{1}{r^2}\Bigl(\pa_r(r^2-2\Bm_0r+\BQ_0^2)\sfS-2(r-\Bm_0)\sfS\Bigr)=0.\\
\widehat{\Box_{g}}(0)^*((r-\Bm_0)H(r-\ehRN)\sfS)&=H(r-\ehRN)\widehat{\Box_{g}}(0)((r-\Bm_0)\sfS)+2\weight\delta(r-\ehRN)\sfS\\
\end{split}
\end{equation}

For the general RN metric $g_b$ with parameter $b=(\Bm, 0,\BQ)$, we notice that the expression of $g_b$ is the same as that of $g$ with $(\Bm_0, 0,\BQ_0)$ replaced by $(\Bm, 0,\BQ)$. Therefore, the above calculation still follows and gives the explicit form of $u_{(\Bm, 0,\BQ),s_1}(\sfS)$ and $u^*_{(\Bm, 0,\BQ),s_1}(\sfS)$.

§ MODE ANALYSIS OF THE $1$-FORM WAVE-TYPE OPERATOR

In this section we shall analyze the modes of the gauge propagation operator $\mathcal{P}_{b,\gamma}=\delta_{g_b} G_{g_b}\wt{\delta}^*_{g_b,\gamma}$ and the gauge potential wave operator $\mathcal{W}_{b,\gamma}=\wt{\delta}_{g_b,\gamma} G_{g_b}\delta_{g_b}^*$, both of which are wave-type operators acting on $1$-form. The former one occurs as the (Einstein part) of gauge propagation operator acting on the gauge $1$-form $D_{g_b}\widetilde{\Upsilon}^E(\dg;g_b)=\tilde{\delta}_{g_b,\gamma}G_{g_b}\dg$, while the latter one acts on the gauge potentials (which generate the pure gauge solutions). We also note that from this section on, the smallness of the charge $\BQ$ is required, i.e., we are restricted in the weakly charged RN and KN (slowly rotating) metric. Throughout this section, the bundle we mostly deal with is scattering $1$-form, so we shall drop “$\scform$” in $\eHb^{s,\ell}(\CX;\scform), \sHb^{s,\ell}(\CX;\scform)$ and $\mathcal{A}(\CX,\scform)$ when it is clear from the context.

Let $B\in C^\infty(M;\mbox{Hom}(T^*M, S^2T^*M)), F\in C^\infty(M;\mbox{Hom}(S^2T^*M, T^*M))$ and $g$ be a Lorentzian metric. Then we define the modified symmetric gradient $\wt{\delta}^*_{g, B}$ and modified negative divergence $\wt{\delta}_{g,F}$ as follows

\begin{equation}\label{eq:defofmodideri}
\wt{\delta}^*_{g, B}=\delta_g^*+B_g,\quad \wt{\delta}_{g,F}=\delta_g+F_g.
\end{equation}

In this manuscript, we will use $B_g, F_g$ of the following form

\begin{gather}\label{eq:defofB}
B_g=B(g;\mathfrak{c},\gamma):=2\gamma\mathfrak{c}\otimes_s(\bullet)-\frac{1}{2}\gamma g^{-1}(\mathfrak{c},\bullet)g,\\\label{eq:defofF}
\end{gather}

where $\gamma\in\BR$, $\mathfrak{c}\in C^\infty_c(X, \scform)$ are stationary $1$-forms with compact support and $\iota_\mathfrak{c}^g(h)=g^{\mu\nu}\mathfrak{c}_\nu h_{\mu\al}$. With the above choices of $B_g$ and $F_g$ in (<ref>) and (<ref>), we find that $\wt{\delta}^*_{g,B}$ and $\wt{\delta}_{g,F}$ are formally adjoint to each other. Throughout this section, we denote

\begin{equation}\label{eq:modifieddelta}
\wt{\delta}^*_{g,\gamma}:=\wt{\delta}^*_{g,B}=\delta_g^*+B_g,\quad \wt{\delta}_{g,\gamma}:=\wt{\delta}_{g,F}=\delta_g+F_g.
\end{equation}

We define the gauge $1$-form

\begin{equation}\label{eq:nonlineargaugeE}
\Upsilon^E(g;g^0)=g(g^0)^{-1}\delta_gG_gg^0
\end{equation}

where $g^0$ is a fixed background metric. In local coordinates, we can write

\[	\Upsilon^E(g;g^0)_\mu=g_{\mu\nu}(g^{-1})^{\kappa\lambda}(\Gamma(g)_{\kappa\lambda}^\nu-\Gamma(g^0)_{\kappa\lambda}^\nu).
\]

We also define the gauge function

\begin{equation}\label{eq:nonlineargaugeM}
\Upsilon^M (g,A;g^0)=\tr_g\delta^*_{g^0}A.
\end{equation}

Then the modified gauge-fixed Einstein-Maxwell system is given by $P_{(g,A),\gamma}=(P^E_{(g,A),\gamma},\ 2P^M_{(g,A),\gamma})=0$ where

\begin{gather}\label{eq:gaugefixedEinstein}
P^E_{(g,A),\gamma}=\Ric(g)-2T(g, dA)-\wt{\delta}^*_{g, \gamma}\Bigl(\Upsilon^E(g;g^0)-\theta(g;g^0)\Bigr)=0,\quad \theta(g;g^0)=F(g^0;\mathfrak{c},\gamma)[G_{g^0} g]\\\label{eq:gaugefixedMaxwell}
P^M_{(g,A),\gamma}=\delta_g dA-d\Bigl(\Upsilon^M(g,A;g^0)-\Upsilon^M(g,A^0;g^0)\Bigr)=0
\end{gather}

where $g^0,A^0$ are fixed background metric and electromagnetic $4$-potential, respectively, and $\theta(g;g^0)$ is linear in $g$ satisfying $\theta(g^0;g^0)=0$.

Fixing $(g^0,A^0)=(g_b,A_b)$ with $b=(\Bm,\Ba,\BQ)$ to be KN metric and electromagnetic $4$-potential, and linearizing (<ref>) and (<ref>) around $(g_b, A_b)$ give

We first recall the  linearized gauge-fixed Einstein-Maxwell system
\[L_{b,\gamma}=(2L^E_{b,\gamma}(\dg,\dA),\ L^M_{b,\gamma}(\dg,\dA))=0
\]

\begin{gather}\label{eq:modifiedlinearizedEinstein}
L^E_{b,\gamma}(\dg,\dA)=D_{g_b}\Ric(\dg)-2D_{(g_b, dA_b)}T(\dg,d\dA)+\wt{\delta}^*_{g_b,\gamma}\wt{\delta}_{g_b,\gamma}G_{g_b}\dg=0,\\\label{eq:modifiedlinearizedMaxwell}
L^M_{b,\gamma}(\dg,\dA)=D_{(g_b, A_b)}(\delta_gdA)(\dg, \dA)+d\delta_{g_b}\dA=0.
\end{gather}

We note that for $(g_b, A_b)$ satisfying the Einstein-Maxwell system, $(2\delta_{g_b}^*\omega,\ \mathcal{L}_{\omega^\sharp}A_b+d\phi)$ solves the linearized gauge-fixed Einstein-Maxwell system (<ref>) and (<ref>) provided that the $1$-form $\omega$ and the scalar function $\phi$ satisfy

\begin{equation}\label{eq:eqnforgaugepotential}
\wt{\delta}_{g_b,\gamma}G_{g_b}\delta^*_{g_b}\omega=0,\quad \delta_{g_b}d\phi=-\delta_{g_b}(\mathcal{L}_{\omega^\sharp}A_b).
\end{equation}

Therefore, zero mode solutions to the above wave-type system of equations (<ref>) give rise to pure gauge zero mode solutions to $L_{(g_b,A_b),\gamma}$.

Dually,  using (<ref>), (<ref>), (<ref>) and the following calculation for $	(D_{g_b}\Ric)^*:\scsym\to\scsym$, $(D_{(g_b,dA_b)}T)^*:\scsym\to\scform\oplus\scsym$ and $(D_{(g_b, A_b)}(\delta_gdA))^*:\scform\to\scform\oplus\scsym$

\begin{gather*}
(D_{(g_b,dA_b)}T)^*(\dg)=(-D_{(g_b, A_b)}(\delta_gdA)(\dg,0),\ D_{(g_b,dA_b)}T(\dg,0)-\frac12(g_{b})_{\mu\nu}\dg^{\al\be}F_{\al}^{\ \gamma}F_{\be\gamma}+\frac12\tr_{g_b}\dg F_{\mu\al}F_{\nu}^{\ \al}),\\
=(-D_{(g_b, A_b)}(\delta_gdA)(G_{g_b}\dg,0),\ G_{g_b}D_{(g_b,dA_b)}T(G_{g_b}\dg,0))\\
(D_{(g_b, A_b)}(\delta_gdA))^*(\dA)=\Bigl(\delta_{g_b}d\dA,\ -D_{(g_b,dA_b)}T(0,d\dA)\Bigr)=\Bigl(d\delta_{g_b}\dA,\ -G_{g_b}D_{(g_b,dA_b)}T(0,d\dA)\Bigr),
\end{gather*}

we conclude

\[L^*_{(g_b,A_b),\gamma}(\dg, \dA)=(2L^{*E}_{(g_b,A_b),\gamma}(\dg,\dA),\  4L^{*M}_{(g_b,A_b),\gamma}(\dg,\dA))
\]

\begin{gather}\label{eq:modifiedlinearizedEinsteinDual}
L^{*E}_{(g_b,A_b),\gamma}(\dg,\dA)=G_{g_b}\Bigl(D_{g_b}\Ric-2D_{(g_b, dA_b)}T+\wt{\delta}^*_{g_b,\gamma}\wt{\delta}_{g_b,\gamma}G_{g_b}\Bigr)(G_{g_b}\dg,\frac14d\dA),\\\label{eq:modifiedlinearizedMaxwellDual}
L^{*M}_{(g_b,A_b),\gamma}(\dg,\dA)=\Bigl(D_{(g_b, A_b)}(\delta_gdA)+d\delta_{g_b}\Bigr)(G_{g_b}\dg, \frac14\dA).
\end{gather}

Therefore, we have $L^*_{(g_b, A_b),\gamma}(2G_{g_b}\delta_{g_b}^*\omega^*, 4\mathcal{L}_{\omega^{*\sharp}}A_b+d\phi^*)=0$ provided that

\begin{equation}\label{eq:eqfordualpotential}
\wt{\delta}_{g_b,\gamma}G_{g_b}\delta_{g_b}^*\omega^*=0,\quad \delta_{g_b}d\phi^*=-4\delta_{g_b}(\mathcal{L}_{\omega^{*\sharp}}A_b).
\end{equation}

Such dual pure gauge zero modes $(2G_{g_b}\delta_{g_b}^*\omega^*, 4\mathcal{L}_{\omega^{*\sharp}}A_b+d\phi^*)$ give rise to zero mode solutions to
\[L^*_{(g_b, A_b),\gamma}(\dg,\dA)=0.
\]

The reason we introduce the constraint damping, i.e., replace $\delta_{g_b}^*$ by $\delta_{g_b, \gamma}^*$ in this manuscript, is that we need to use the invertibility of the corresponding gauge propagation operator $\delta_{g_b}G_{g_b}\tilde{\delta}^*_{g_b}$ to exclude the generalized modes, which grows linearly in $t_{b,*}$ and whose leading term is given by linearized (in $(\dot{\Bm}, 0,\dot{\BQ})$) KN solutions, to the linearized gauge-fixed Einstein-Maxwell equations. We also note that we use a perturbation argument to
prove the invertibility of the gauge propagation operator $\delta_{g_b}G_{g_b}\tilde{\delta}^*_{g_b}$ on the weakly charged and slowly rotating KN metrics $g_b$. Moreover, we use the generalized linearized gauge condition for the Einstein part $\wt{\delta}_{g_b, \gamma}G_{g_b}\dg=0$ in this paper to exclude the zero mode of no geometric significance, see <cit.>.

Let $\tilde{\chi}(r)\in C_c^\infty((r_-, 3\Bm_0))$ such that $\tilde{\chi}=1$ near $r=2\Bm_0$ (and thus near the event horizon of $g_b$ with $b$ near $(\Bm_0, 0,\BQ_0)$ where $\abs{\BQ_0}\ll\Bm_0$). We use the following

\begin{equation}\label{eq:defofc}
\mathfrak{c}=\tilde{\chi}(r)(dt_0-\mathfrak{b}dr), \quad t_0=t+r_{(\Bm_0, 0,\BQ_0),*}
\end{equation}

with $\mathfrak{b}$ to be determined later, to define $\wt{\delta}^*_{g_b, \gamma}, \wt{\delta}_{g_b, \gamma}, L_{(g_b, A_b),\gamma}, L^*_{(g_b, A_b),\gamma}$. As discussed in (<ref>), we then define gauge propagation operator

\begin{equation}
\mathcal{P}_{b,\gamma}:=\delta_{g_b}G_{g_b}\wt{\delta}_{g_b, \gamma}^*
\end{equation}

and gauge potential wave operator

\begin{equation}
\mathcal{W}_{b,\gamma}:=\wt{\delta}_{g_b,\gamma}G_{g_b}\delta_{g_b}^*.
\end{equation}

We notice that $\mathcal{P}_{b,\gamma}$ and $\mathcal{W}_{b,\gamma}$ are formally dual to each other. However, on the level of function spaces, they are not because we need to work with the extendible $b$-Sobolev spaces for both when analyzing the modes of $L_{(g_b, A_b),\gamma}$. We also note that the dual pure gauge zero mode solutions to $L^*_{(g_b, A_b),\gamma}(\dg,\dA)=0$ (as discussed in (<ref>)) are closely related to the zero mode solutions to $\mathcal{W}_{b,\gamma}\omega^*=0$.

In order to study the modes of $\mathcal{P}_{b,\gamma}$ and $\mathcal{W}_{b, \gamma}$ for $b=(\Bm,\Ba,\BQ)$ near $b_0=(\Bm_0,0,\BQ_0)$, we define

\begin{equation}\label{eq:fourierCPW}
\widehat{\mathcal{P}_{b,\gamma}}(\sigma):=e^{i\sigma t_{b,*}}\mathcal{P}_{b,\gamma}e^{-i\sigma t_{b,*}},\quad \widehat{\mathcal{W}_{b,\gamma}}(\sigma):=e^{i\sigma t_{b,*}}\mathcal{W}_{b,\gamma}e^{-i\sigma t_{b,*}}
\end{equation}

where $t_{b,*}=(t+r_{b_0,*})\chi_0(r)+(t-r_{(\Bm,0,\BQ),*})(1-\chi_0(r))$ is defined as in (<ref>).

We first record the following result about Schwarzschild metric which we rely on in the analysis of $\widehat{\mathcal{P}_{b,\gamma}}(\sigma)$ and $\widehat{\mathcal{W}_{b, \gamma}}(\sigma)$ on weakly charged and slowly rotating KN metrics.

Let $(g,M_{(\Bm_0,0,0)})$ be a Schwarzschild spacetime with mass $\Bm_0$ and $\mathcal{P}_0=\delta_gG_g\delta_{g}^*$ be the wave operator acting on $1$-form. Define $\widehat{\mathcal{P}_0}(\sigma):=e^{i\sigma t_{0,*}}\mathcal{P}_0e^{-i\sigma t_{0,*}}$ with $t_{0,*}=(t+r_{0,*})\chi_0(r)+(t-r_{0,*})(1-\chi_0(r))$ and $ r_{0,*}=r+2\Bm_0\log(r-2\Bm_0)$. Then we have

* If $\IM \sigma\geq0$ and $\sigma\neq 0$, then

\begin{equation}
\widehat{\mathcal{P}_0}(\sigma):\{\omega\in\eHb^{s, \ell}(\CX;\scform): \widehat{\mathcal{P}_0}(\sigma)\omega\in\eHb^{s,\ell+1}(\CX; \scform)\}\to \eHb^{s,\ell+1}(\CX;\scform)
\end{equation}

is invertible for $s>\frac 32, \ell<-\frac12, s+\ell>-\frac 12$.

* If $s>\frac 32$ and $-\frac 32<\ell<-\frac 12$, then stationary operator

\begin{equation}
\widehat{\mathcal{P}_0}(0):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{P}_0}(0)\omega\in\eHb^{s-1,\ell+2}(\CX; \scform)\}\to \eHb^{s-1,\ell+2}(\CX;\scform)
\end{equation}

has $1$-dimensional kernel and cokernel.

More explicitly, with $v=t+r_{0,*}=t+r+2\Bm_0\log(r-2\Bm_0)$ we have

\begin{equation}\label{eq:g1formmodeSchw}
\begin{split}
\ker\widehat{\mathcal{P}_0}(0)\cap \eHb^{\infty, -1/2-}(\CX; \scform)&=\langle\omega_{ s_0}\rangle,\quad 	\omega_{ s_0}=r^{-1}(dv-dr)\\
\ker\widehat{\mathcal{P}_0}(0)^*\cap \sHb^{-\infty, -1/2-}(\CX; \scform)&=\langle\omega^*_{ s_0}\rangle,\quad \omega^*_{ s_0}=\delta(r-2\Bm_0)dr
\end{split}.
\end{equation}

According to (<ref>), $\widehat{\mathcal{P}_0}(0)$ restricted to $1$-forms of scalar type $l=1$ (and vector type $l=1$ respectively)
is invertible. Since  $\widehat{\mathcal{W}_{b_0,\gamma}}(0)$ also maps $1$-forms of scalar type $l=1$ (and vector type $l=1$ respectively) to the same type and is a small perturbation of $\widehat{P_0}(0)$ for  $b_0=(\Bm_0, 0, \BQ_0)$ with $\abs{\BQ_0}\ll\Bm_0$ and $\gamma\ll1$, it follows that $\widehat{\mathcal{W}_{b_0,\gamma}}(0)^{-1}$ restricted to $1$-forms of scalar type $l=1$ (and $1$-forms of vector type $l=1$ respectively) exists with norm $\mathcal{O}(1)$.

Next we will show that $\widehat{\mathcal{P}_{b,\gamma}}(\sigma)$ and $\widehat{\mathcal{W}_{b, \gamma}}(\sigma)$ have index $0$.

For $b=(\Bm, \Ba, \BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$ and $\gamma\in\BR$ sufficiently small, we let $\widehat{\mathcal{P}_{b,\gamma}}(\sigma)$ be defined as in (<ref>). Then the operators

\begin{equation}\label{eq:indexofPnonozero}
\widehat{\mathcal{P}_{b,\gamma}}(\sigma):\{\omega\in\eHb^{s, \ell}(\CX;\scform): \widehat{\mathcal{P}_{b,\gamma}}(\sigma)\omega\in\eHb^{s,\ell+1}(\CX; \scform)\}\to \eHb^{s,\ell+1}(\CX;\scform)
\end{equation}

with $\IM\sigma\geq 0, \sigma\neq 0, s>\frac 32, \ell<-\frac 12, s+\ell>-\frac 12$, and

\begin{equation}\label{eq:indexofP0}
\widehat{\mathcal{P}_{b,\gamma}}(0):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{P}_{b,\gamma}}(0)\omega\in\eHb^{s-1,\ell+2}(\CX; \scform)\}\to \eHb^{s-1,\ell+2}(\CX;\scform)
\end{equation}

with $s>\frac 32, -\frac 32<\ell<-\frac 12$, have index $0$. The above result also holds for $\widehat{\mathcal{W}_{b,\gamma}}(\sigma)$.

To be completed We first analyze the case $\sigma=0$. We shall use a deformation argument to reduce the index of $\widehat{P}_{b,\gamma}(0)$ to that of a $4\times4$ diagonal matrix of scalar wave operators $\widehat{\Box_{g_b}}(0)$. More specifically,

As for the case $\sigma\neq 0$, we proceed in the same way as in the fifth step in the proof of Theorem <ref>, that is, we exploit the high energy estimates, uniform Fredholm estimates for $\widehat{\mathcal{P}_{b,\gamma}}(\sigma)$ in  Theorem and then a perturbation argument to finish the proof for (<ref>).

The proof for $\widehat{\mathcal{W}_{b,\gamma}}(\sigma)$ is completely analogous.

Inspecting the last step in the proof of Theorem <ref>, we find that
for $s>\frac 12,\ell\in (-\frac 32, \frac 12)$ and $\IM\sigma\geq 0,\abs{\sigma}<c, \abs{b-b_0}<\epsilon_2$ for some $c, \epsilon_2>0$,

\[\widehat{\Box_{g_{b}}}(\sigma):\{u\in\eHb^{s,\ell}(\CX):\widehat{\Box_{g_{b}}}(\sigma)u\in\eHb^{s-1,\ell+2}(\CX)\}\to \eHb^{s-1,
\ell+2}(\CX)
\]

is invertible. Then a deformation argument as above implies for the same given $s,\ell,\sigma,b$

\[\widehat{\mathcal{P}_{b,\gamma}}(\sigma):\{\omega\in\eHb^{s,\ell}(\CX;\scform):\widehat{\mathcal{P}_{b,\gamma}}(\sigma)u\in\eHb^{s-1,\ell+2}(\CX;\scform)\}\to \eHb^{s-1,
\ell+2}(\CX;\scform)
\]

has index $0$. The same result holds for $\widehat{\mathcal{W}_{b,\gamma}}(\sigma)$ as well.

§.§ Mode stability of the gauge propagation operator

We first record the following two propositions about Schwarzschild metric which are needed in the course of the proof of mode stability of the gauge propagation operator $\mathcal{P}_{b,\gamma}$ for the weakly charged RN metric and also slowly rotating and weakly charged KN metric.

Let $(g,M_{(\Bm_0,0,0)})$ be the Schwarzschild spacetime with mass $\Bm_0$ and $\mathcal{P}_{0,\gamma}=\delta_gG_g\tilde{\delta}_{g,\gamma}^*$ where $\tilde{\delta}^*_{g,\gamma}$ is defined as in (<ref>) and (<ref>) with $\mathfrak{c}=\tilde{\chi}(r)(dv-\mathfrak{b}dr), v=t+r_{0,*}=t+r+2\Bm_0\log(r-2\Bm_0)$ and $\mathfrak{b}\neq 1$. Then there exists $\gamma_0>0$ such that for fixed $\gamma$ with $0<\abs{\gamma}<\gamma_0$, the following holds for $\widehat{\mathcal{P}_{0,\gamma}}(\sigma):=e^{i\sigma t_{0,*}}\mathcal{P}_{0,\gamma}e^{-i\sigma t_{0,*}}$ with $t_{0,*}=(t+r_{0,*})\chi_0(r)+(t-r_{0,*})(1-\chi_0(r))$.

\begin{equation}
\widehat{\mathcal{P}_{0,\gamma}}(0):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{P}_{0,\gamma}}(0)\omega\in\eHb^{s-1,\ell+2}(\CX; \scform)\}\to \eHb^{s-1,\ell+2}(\CX;\scform)
\end{equation}

with $s>2, -\frac 32<\ell<-\frac 12$ is invertible.

Let $\widehat{\mathcal{P}_{0,\gamma}}(\sigma)$ be defined as in Proposition <ref> with $\mathfrak{b}>1$, then we have for $\IM\sigma\geq 0, \sigma\neq 0$ and $s>2, \ell<-\frac 12, s+\ell>-\frac 12$

\begin{equation}
\widehat{\mathcal{P}_{0,\gamma}}(\sigma):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{P}_{0,\gamma}}(\sigma)\omega\in\eHb^{s,\ell+1}(\CX; \scform)\}\to \eHb^{s,\ell+1}(\CX;\scform)
\end{equation}

is invertible.

We note that $\delta^*_{g,\gamma}$ defined in <cit.> is slightly different from that in this manuscript ((<ref>) and (<ref>)) in the way that the coefficient of the second term in $B$ in [57] (which is denoted by $E$ there) is $-1$ instead of $-1/2$. However, this difference has no influence on the conclusions stated in the above two propositions. In particular, it does not affect the calculation in <cit.>.

Next we shall prove the analogues of the above two propositions for RN and slowly rotating KN metric with small charge by a perturbation argument.

We define $\mathfrak{c}$ as in (<ref>) with $\mathfrak{b}>1$. With $\gamma_0$ as in Propositions <ref> and <ref>, we fix $\gamma\in(0,\gamma_0)$.

Let $b_0=(\Bm_0, 0,\BQ_0)$ and $\mathfrak{b}>1$. Fix $\gamma\in(0,\gamma_0)$, there exists a small constant $C(\gamma)>0$ such that for the weakly charged RN metric $g_{b_0}$ with $\abs{\BQ_0}<C(\gamma)$, the following holds.

* If $\IM \sigma\geq0$ and $\sigma\neq 0$,

\begin{equation}\label{eq:CPmodenonzero}
\widehat{\mathcal{P}_{b_0,\gamma}}(\sigma):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{P}_{b_0,\gamma}}(\sigma)\omega\in\eHb^{s,\ell+1}(\CX; \scform)\}\to \eHb^{s,\ell+1}(\CX;\scform)
\end{equation}

is invertible when $s>2, \ell<-\frac12, s+\ell>-\frac 12$.

* If $s>2$ and $-\frac 32<\ell<-\frac 12$, then stationary operator

\begin{equation}\label{eq:CPmodezero}
\widehat{\mathcal{P}_{b_0,\gamma}}(0):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{P}_{b_0,\gamma}}(0)\omega\in\eHb^{s-1,\ell+2}(\CX; \scform)\}\to \eHb^{s-1,\ell+2}(\CX;\scform)
\end{equation}

is invertible.

Both statements also hold for the weakly charged and slowly rotating KN metric $g_b$ with $b=(\Bm, \Ba,\BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$.

We note that a perturbation argument as in the final step in the proof of Theorem <ref> allows us to extend Propositions <ref> and <ref> to weakly charged RN metrics, which however is a smooth family of Lorentzian metric on the fixed manifold $M_{(\Bm_0, 0,0)}$ and is a pull back of the RN metric $g_{b_0}$ defined in (<ref>). We need to prove that Propositions <ref> and <ref> in fact hold for $g_{b_0}$ with sufficiently small charge as well. Specifically, let $(g, M_{(\Bm_0, 0,0)})$ be the Schwarzschild spacetime

\[
g=-(1-\frac{2\Bm_0}{r})dv^2+2dv dr+r^2\sg,\quad
M_{(\Bm_0, 0,0)}=\BR_{v}\times[r_-, \infty)\times\BS^2
\]

where $r_-\in(0,2\Bm_0)$ is a fixed number and $v=t+r_{0,*}=t+r+2\Bm_0\log(r-2\Bm_0)$. We introduce another coordinates for $(g, M_{(\Bm_0, 0,0)})$: $(\tilde{t}_{\chi_0}, r,\theta,\phi)$ where $\tilde{t}_{\chi_0}=t+\int_{4\Bm_0}^r s^{-2}(s^2-2\Bm_0 s)\chi_0(s)\,ds$. Let $g_{b_0}$ be the RN metric as defined in (<ref>). We then define $\tilde{g}_{b_0}=(\Psi_{b_0})^*g_{b_0}$ as the pull back of $g_{b_0}$ under a diffeomorphism $\Psi_{b_0}$

\[
\Psi_{b_0}: M_{(\Bm, 0,0)}\to M, \quad (t_{\chi_0}, r, \theta ,\phi)(\Psi_{b_0}(p))=(\tilde{t}_{\chi_0}, r,\theta, \phi)(p).
\]

Then we have

\begin{align}\label{eq:defofg_b}
g_{b_0}&=-\weight dt_{\chi_0}^2+2\chi_0 dt_{\chi_0} dr+\frac{1-\chi_0^2}{\weight}dr^2+r^2\sg,\\\label{eq:defoftildeg_b}
\tilde{g}_{b_0}&=-\weight d\tilde{t}_{\chi_0}^2+2\chi_0 d\tilde{t}_{\chi_0} dr+\frac{1-\chi_0^2}{\weight}dr^2+r^2\sg.
\end{align}

We now define

\[
\mathcal{P}_{\tilde{g}_{b_0},\gamma}:=\delta_{\tilde{g}_{b_0}}G_{\tilde{g}_{b_0}}\delta^*_{\tilde{g}_{b_0},\gamma}
\]

with $\mathfrak{c}=\tilde{\chi}(r)(dv-\mathfrak{b}dr)$ and $v=t+r_*$ in the definition of $\delta^*_{\tilde{g}_{b_0},\gamma}$. In view of (<ref>) and (<ref>), we find that $\mathcal{P}_{b_0,\gamma}$ can be obtained by replacing $\pa_{\tilde{t}_\chi}$ in $\mathcal{P}_{\tilde{g}_{b_0},\gamma}$ by $\pa_{t_\chi}$.

\begin{align*}
\widehat{\mathcal{P}_{b_0,\gamma}}(\sigma)&=e^{i\sigma t_{b_0,*}}\mathcal{P}_{b_0,\gamma}e^{-i\sigma t_{b_0,*}},\quad t_{b_0,*}=(t+r_{b_0,*})\chi_0(r)+(t-r_{b_0,*})(1-\chi_0(r)), \\
\widehat{\mathcal{P}_{\tilde{g}_{b_0},\gamma}}(\sigma)&=e^{i\sigma \tilde{t}_{b_0,*}}\mathcal{P}_{\tilde{g}_{b_0},\gamma}e^{-i\sigma \tilde{t}_{b_0,*}},\quad \tilde{t}_{b_0,*}=(t+r_{0,*})\chi_0(r)+(t-r_{b_0,*})(1-\chi_0(r)),
\end{align*}

we rewrite

\begin{align*}
\widehat{\mathcal{P}_{b_0,\gamma}}(\sigma)&=	e^{i\sigma (t_{\chi_0}+F_1(r))}\mathcal{P}_{b_0,\gamma}e^{-i\sigma (t_{\chi_0}+F_1(r))},\\
\widehat{\mathcal{P}_{\tilde{g}_{b_0},\gamma}}(\sigma)&=	e^{i\sigma (\tilde{t}_{\chi_0}+F_2(r))}\mathcal{P}_{\tilde{g}_{b_0},\gamma}e^{-i\sigma (\tilde{t}_{\chi_0}+F_2(r))}.
\end{align*}

\[
\widehat{\mathcal{P}_{b_0,\gamma}}(\sigma)=e^{i\sigma(F_1(r)-F_2(r))}\widehat{\mathcal{P}_{\tilde{g}_{b_0},\gamma}}(\sigma)e^{-i\sigma(F_1(r)-F_2(r))}
\]

where $F_1(r)-F_2(r)\in C_c^\infty((2\Bm_0, \infty))$. As discussed at the beginning, Propositions <ref> and <ref> hold for $\widehat{\mathcal{P}_{\tilde{g}_{b_0},\gamma}}(\sigma)$, and thus for $\widehat{\mathcal{P}_{b_0,\gamma}}(\sigma)$ as well with $\abs{\BQ_0}<C(\gamma)$ where $C(\gamma)$ is a sufficiently small constant.

Finally  it remains to analyze the weakly charged and slowly rotating KN case.  It follows from a perturbation argument as in the final step in the proof of Theorem <ref>.

§.§ Mode stability of the gauge potential wave operator

Now we turn to the analysis of the gauge potential wave operator $\mathcal{W}_{b,\gamma}=\wt{\delta}_{g_b,\gamma}G_{g_b}\delta_{g_b}^*$. We will first prove the analogues of Propositions <ref> and <ref> for $\mathcal{W}_{0,\gamma}$. We note that the proofs closely follow <cit.>. However, we present the proofs in detail for completeness here. Then proceeding as in Theorem <ref>, we can obtain its counterpart for $\mathcal{W}_{b,\gamma}$.

Let $(g,M_{(\Bm_0,0,0)})$ be the Schwarzschild spacetime with mass $\Bm_0$ and $\mathcal{W}_{0,\gamma}=\tilde{\delta}_{g,\gamma}G_g\delta_{g}^*$ where $\tilde{\delta}_{g,\gamma}$ is defined as in (<ref>) and (<ref>) with $\mathfrak{c}=\tilde{\chi}(r)(dv-\mathfrak{b}dr), v=t+r_{0,*}=t+r+2\Bm_0\log(r-2\Bm_0)$ and $\mathfrak{b}\neq 1$. Then there exists $\gamma_0>0$ such that for fixed $\gamma$ with $0<\abs{\gamma}<\gamma_0$, the following holds for $\widehat{\mathcal{W}_{0,\gamma}}(\sigma):=e^{i\sigma t_{0,*}}\mathcal{W}_{0,\gamma}e^{-i\sigma t_{0,*}}$ with $t_{0,*}=(t+r_{0,*})\chi_0(r)+(t-r_{0,*})(1-\chi_0(r))$.

\begin{equation}
\widehat{\mathcal{W}_{0,\gamma}}(0):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{W}_{0,\gamma}}(0)\omega\in\eHb^{s-1,\ell+2}(\CX)\}\to \eHb^{s-1,\ell+2}(\CX)
\end{equation}

with $s>2, -\frac 32<\ell<-\frac 12$ is invertible.

Fix $-\frac 32<\ell<-\frac 12, s>2$ and let

\[
\mathcal{X}^{s,\ell}:=\{u\in\eHb^{s,\ell}(\CX):\widehat{\mathcal{W}_{0,\gamma}}(0)u\in\eHb^{s-1,\ell+2}(\CX)\}.
\]

In view of Lemma <ref>, we know that $\widehat{\mathcal{W}_{0,\gamma}}(0):\mathcal{X}^{s,\ell}\to\eHb^{s-1,\ell+2}(\CX)$ is Fredholm of index $0$ when $\gamma$ is sufficiently small. As $\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_2}}(0)$ is a first-order differential operator with compactly supported coefficients, the space $\mathcal{X}^{s,\ell}$ does not depend on $\gamma$.

We split the domain $\mathcal{X}^{s,\ell}$ and the target space $\eHb^{s,\ell}(\CX)$ as follows

\begin{align*}
\mathcal{X}^{s,\ell}&=\mathcal{K}^{\perp}\oplus \mathcal{K},\quad \mathcal{K}=\ker\widehat{\mathcal{P}_0}(0)\cap\mathcal{X}^{s,\ell}=\langle \omega_{s_0}\rangle\\
\eHb^{s,\ell}(\CX)&=\mathcal{R}\oplus \mathcal{R}^\perp,\quad \mathcal{R}=\Ran_{\mathcal{X}^{s,\ell}}\widehat{\mathcal{P}_0}(0)=\mbox{ann }\omega^*_{s_0}
\end{align*}

where ann means annihilator, $\mathcal{K}^\perp$ is a complementary subspace to $\mathcal{K}$ inside $\mathcal{X}^{s,\ell}$ while $\mathcal{R}^\perp=\langle \eta\rangle$ is complementary to $\mathcal{R}$ inside $\eHb^{s-1,\ell+2}$ (in other word, $\mathcal{R}^\perp$ is the cokernel of $\widehat{\mathcal{P}_0}(0)$), and $ \omega_0, \omega_0^*$ are defined as in (<ref>). We may choose $\eta\in C_c^\infty(X;\scform)$ satisfying $\angles{\eta}{\omega_{s_0}^*}=1$. Then the operator $\widehat{\mathcal{W}_0}(0)$ can be rewritten as

\begin{equation}\label{decomofW}
\widehat{\mathcal{W}_{0,\gamma}}(0)=\begin{pmatrix}
\mathcal{W}_{00}+\gamma\mathcal{W}^\flat_{00}&\gamma\mathcal{W}^\flat_{01}\\
\gamma\mathcal{W}^\flat_{10}&\gamma\mathcal{W}^\flat_{11}
\end{pmatrix}.
\end{equation}

We notice that $\mathcal{W}_{00}=\widehat{\mathcal{P}_0}(0)|_{\mathcal{K}^\perp}:\mathcal{K}^\perp\to \mathcal{R}$ is invertible, and so is $\mathcal{W}_{00}+\gamma\mathcal{W}^\flat_{00}$ provided that $\gamma$ is small enough. Moreover, if we identify $\mathcal{K}\cong\BC$ via $c\omega_{s_0}\mapsto c$ and $\mathcal{R}^\perp\cong\BC$ via $c\eta\mapsto \angles{c\eta}{\omega_{s_0}^*}=c$, correspondingly, $\mathcal{W}^\flat_{11}$ can be identified as an endmorphism of $\BC$, and thus is simply a number which can be computed as follows.

\begin{equation}\label{eq:calofW_11flat}
\begin{split}
\mathcal{W}_{11}^\flat&=\gamma^{-1}\angles{\bigl(\widehat{\mathcal{W}_{0,\gamma}}(0)-\widehat{\mathcal{P}_{0}}(0)\bigr)\omega_{s_0}}{\omega_{s_0}^*}\\
&=\angles{\delta_g^*\omega_{s_0}}{\big(2\mathfrak{c}\otimes_s\omega_{s_0}^*-\frac 12 g^{-1}(\mathfrak{c},\omega_{s_0}^*)g\big)}\\
\end{split}
\end{equation}

where we use $g=-(1-\frac{2\Bm_0}{r})dv^2+2dvdr+r^2\sg$, $2\mathfrak{c}\otimes_s\omega^*_{s_0}=\delta(r-2\Bm_0)\tilde{\chi}(r)(2dvdr-2\mathfrak{b}dr^2)$, $g^{-1}(\mathfrak{c},\omega_{s_0}^*)=\delta(r-2\Bm_0)\tilde{\chi}(r)(1-\mathfrak{b}(1-\frac{2\Bm_0}{r}))$ and $\delta_g^*\omega_{s_0}=-\frac{2\Bm_0^2}{r^4}dv^2-2(\frac{1}{2r^2}+\frac{\Bm_0}{r^3})dvdr+\frac{1}{r^2}dr^2-(1-\frac{2\Bm_0}{r})\sg$ in the last step.

Now suppose $(\omega_0,\omega_1)\in(\mathcal{K}^\perp\oplus\mathcal{K})\cap\ker	\widehat{\mathcal{W}_{0,\gamma}}(0)$, then we have

\[
\omega_0=-\gamma(\mathcal{W}_{00}+\gamma\mathcal{W}_{00}^\flat)^{-1}\mathcal{W}_{01}^\flat\omega_1
\]

and  thus

\[
\Big(\mathcal{W}_{11}^\flat-\gamma\mathcal{W}_{10}^\flat(\mathcal{W}_{00}+\gamma\mathcal{W}_{00}^\flat)^{-1}\mathcal{W}_{01}^\flat\Big)\omega_1=0.
\]

if $\gamma$ is nonzero. When $\mathfrak{b}\neq 1$, $\mathcal{W}_{11}^\flat$ is invertible, and thus so is $\mathcal{W}_{11}^\flat-\gamma\mathcal{W}_{10}^\flat(\mathcal{W}_{00}+\gamma\mathcal{W}_{00}^\flat)^{-1}\mathcal{W}_{01}^\flat$ if $\gamma$ is sufficiently small. So we must have $\omega_1=0$ and thus $\omega_0=0$. This proves the injectivity of $\widehat{\mathcal{W}_{0,\gamma}}(0)$. The surjectivity follows from the fact that $\widehat{\mathcal{W}_{0,\gamma}}(0)$ is Fredholm of index $0$.

As explained in (<ref>) and (<ref>), $\mathcal{W}_{b,\gamma}$ acts as the gauge potential wave operator. In order to obtain the mode stability of $L_{g_b,\gamma}(\dg,\dA)=0$ for $\IM\sigma\geq 0, \sigma\neq 0$, we need to  prove that for the operator $\mathcal{W}_{b,\gamma}$. In other words, we need to show that for suitable choices of $\mathfrak{b}$ and small $\gamma$, $\mathcal{W}_{b,\gamma}$ has no modes with $\sigma\neq 0$ in the closed upper half plane. We first prove this for Schwarzschild metric, and then the RN and slowly rotating KN metric with small charge follows from a perturbation argument as in Theorem <ref>.

Let $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ be defined as in Proposition <ref> with $\mathfrak{b}>1$, then there exists $\tilde{\gamma}_0>0$ such that for $0<\gamma<\tilde{\gamma}_0$, $\IM\sigma\geq 0, \sigma\neq 0$ and $s>2, \ell<-\frac 12, s+\ell>-\frac 12$

\begin{equation}\label{eq:WnonzeroinSchw}
\widehat{\mathcal{W}_{0,\gamma}}(\sigma):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\omega\in\eHb^{s,\ell+1}(\CX; \scform)\}\to \eHb^{s,\ell+1}(\CX;\scform)
\end{equation}

is invertible.

In the course of the proof of Proposition <ref>, we closely follow the arguments in <cit.>. As explained there, the proof consists of two steps:

* We show in Lemma <ref> that for $\mathfrak{b}>1$ and for all sufficiently small $\gamma$, $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ has no modes in a fixed neighborhood of $\sigma=0$ in the closed upper half plane.

* In the proof of Proposition <ref>, we proceed as in the final step in the proof of Theorem <ref> to prove the mode stability of $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ in the closed upper half plane.

Fix $s>2, -\frac 32<\ell<-\frac12$, the domains

\begin{equation}
\mathcal{X}^{s,\ell}(\sigma):=\{\omega\in\eHb^{s,\ell}(\CX): \widehat{\mathcal{W}_{0,\gamma}}(\sigma)\omega\in\eHb^{s-1,\ell+2}(\CX)\}
\end{equation}

of the operators $\widehat{\mathcal{W}_{0,\gamma}}(\sigma):\mathcal{X}^{s,\ell}(\sigma)\to\eHb^{s-1,\ell+2}$ now depend on $\sigma$ (but still independent of $\gamma$). As a consequence, we need the following technical lemma which allows us to pass to operators with fixed domain and target space.

Let $s>2, -\frac 32<\ell<-\frac12$ and $\mathfrak{b}\neq 1$ and fix $\gamma_1$ satisfying $0<\abs{\gamma_1}<\gamma_0$ where $\gamma_0$ is as in Proposition <ref>. Then $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma):\mathcal{X}^{s,\ell}(\sigma)\to\eHb^{s-1,\ell+2}(\CX)$ is invertible for $\sigma\in\BC$ with $\IM\sigma\geq 0, \abs{\sigma}<c$ where $c$ is some small constant. Moreover, $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ is continuous in $\sigma$ with values in $\mathcal{L}_{\mathrm{weak}}(\eHb^{s-1,\ell+2}(\CX), \eHb^{s,\ell}(\CX))$ (the space of linear bounded operators equipped with weak operator topology), while continuous in $\sigma$ with values in $\mathcal{L}_{\mathrm{op}}(\eHb^{s-1+\epsilon,\ell+2+\epsilon}(\CX), \eHb^{s-\epsilon,\ell-\epsilon}(\CX))$ (norm topology) for any $\epsilon>0$.

By Theorem <ref>, we have the uniform Fredholm estimates for $\omega\in\mathcal{X}^{s,\ell}(\sigma)$

\begin{equation}
\norm{\omega}_{\eHb^{s,\ell}(\CX)}\leq C\Bigl(\norm{\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\omega}_{\eHb^{s-1,\ell+2}(\CX)}+\norm{\omega}_{\eHb^{s_0,\ell_0}(\CX)}\Bigr)
\end{equation}

for $\IM\sigma\geq 0$ with $\abs{\sigma}$ being small and $s_0<s,\ell_0<\ell$. We now follow the arguments in <cit.> to show  that the compact error term $\norm{\omega}_{\eHb^{s_0,\ell_0}(\CX)}$ can be dropped (while we may need to take larger constant $C$) if $\abs{\sigma}$ is sufficiently small. Concretely, suppose for the sake of contradiction that there exists a sequence $\delta_j\to0$ and a sequence $\omega_j$ with $\norm{\omega_j}_{\eHb^{s,\ell}(\CX)}=1$ and $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)\omega_j\in\eHb^{s-1,\ell+2}(\CX)$ such that $1=\norm{\omega_j}_{\eHb^{s,\ell}(\CX)}\geq j\norm{\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)\omega_j}_{\eHb^{s-1,\ell+2}(\CX)}$, then $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)\omega_j\to 0$ in $\eHb^{s-1,\ell+2}(\CX)$ and $\liminf_{j\to\infty}\norm{\omega_j}_{\eHb^{s_0,\ell_0}(\CX)}\geq C^{-1}>0$. As a consequence, there exists a subsequence $\omega_j\to \omega$ weakly in $\eHb^{s, \ell}(\CX)$ and strongly in $\eHb^{s_0, \ell_0}(\CX)$ with the limit $\omega$ being non-zero. Moreover,

\[
\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)\omega_j-\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega=\bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)-\widehat{\mathcal{W}_{0,\gamma_1}}(0)\bigr)\omega_j+\widehat{\mathcal{W}_{0,\gamma_1}}(0)\bigl(\omega_j-\omega\bigr)\to 0\quad\mbox{in}\quad \eHb^{s_0-2, \ell_0}(\CX)
\]

since $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)\to\widehat{\mathcal{W}_{0,\gamma_1}}(0)$ as bounded operators in $\mathcal{L}(\eHb^{s_0, \ell_0}(\CX), \eHb^{s_0-2, \ell_0}(\CX))$ and $\omega_j\to\omega$ in $\eHb^{s_0, \ell_0}(\CX)$. Therefore, $\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega=0$. But this contradicts Proposition <ref>, and thus proves the injectivity of $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)$ for $\sigma$ with small $\abs{\sigma}$ in the closed upper half plane. Since $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)$ has index $0$, it follows that $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ exists with a uniform bound in a small neighborhood of $\sigma=0$ in the closed upper half plane.

Next we prove the sequential continuity of $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ in $\sigma$ (which is equivalent to continuity). Suppose $\delta_j\to\delta$, one need to show $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}\to\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ in weak operator topology of $\mathcal{L}(\eHb^{s-1,\ell+2}(\CX), \eHb^{s,\ell}(\CX))$. Suppose $\{f_j\}$ is a sequence in $\eHb^{s-1,\ell+2}$ which is norm converging to $f$, let $\omega_j=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f_j$ and we shall show that $\omega_j$ converges weakly to $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}f$ in $\eHb^{s,\ell}(\CX)$. In view of the uniform bound of $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}$ proved in the first step above, we see that $\omega_j$ is uniformly bounded in $\eHb^{s,\ell}(\CX)$. Therefore, there exists a subsequence $\omega_{j_k}\to \omega$ weakly in $\eHb^{s,\ell}(\CX)$, then $f_{j_k}= \widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_{j_k})\omega_{j_k}\to \widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\omega$ weakly in $\eHb^{s-2,\ell}(\CX)$. Meanwhile by assumption $f_{j_k}\to f$ strongly in $\eHb^{s-1,\ell+2}(\CX)$ and thus $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\omega=f$. This implies that $\omega$ is uniquely determined by $f$ and independent of the subsequence. Namely, every subsequence of $\omega_j$ has a subsequence converging weakly to the same $\omega$. This means that the full sequence $\omega_j\to\omega=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f$ weakly in $\eHb^{s,\ell}(\CX)$ and thus proves the continuity in the weak operator topology (Otherwise, one can find $g\in\sHb^{-s,-\ell}(\CX)), \delta>0$ and a subsequence $\{\omega_{j_k}\}\subset\{w_j\}$ such that $\abs{g(\omega_{j_k})-g(\omega)}\geq \delta$. However, by assumption $\{\omega_{j_k}\}$ should have a subsequence converging weakly to $\omega$, which is impossible).

As for the continuity in the norm topology, we shall prove it by contradiction. Suppose there exists a $\delta>0$ and a sequence $\sigma_j\to\sigma$ such that $\norm{\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}}_{\mathcal{L}(\eHb^{s-1+\epsilon,\ell+2+\epsilon}(\CX), \eHb^{s-\epsilon,\ell-\epsilon}(\CX))}\geq \delta$, then one can find a bounded sequence $f_j\in\eHb^{s-1+\epsilon,\ell+2+\epsilon}$ such that

\begin{equation}\label{eq:nonormconti}
\norm{\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f_j-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}f_j}_{ \eHb^{s-\epsilon,\ell-\epsilon}(\CX)}\geq \delta.
\end{equation}

Passing to a subsequence, we may assume that $f_j\to f$ weakly in $\eHb^{s-1+\epsilon,\ell+2+\epsilon}(\CX)$ and strongly in $\eHb^{s-1,\ell+2}(\CX)$. Then it follows that  $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}f_j\to \widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}f$ in $\eHb^{s,\ell}(\CX)$. However, by the continuity in the weak operator topology, we see that $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f_j\to \widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}f$ weakly in $\eHb^{s,\ell}(\CX)$ and thus (passing to a subsequence) strongly in $\eHb^{s-\epsilon,\ell-\epsilon}(\CX)$, which contradicts (<ref>). This finishes the proof of the continuity in the norm topology.

With the help of the invertibility of $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)$, the analysis of $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ can be converted to that of another operator $ \widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}:\eHb^{s-1,\ell+2}(\CX)\to\eHb^{s-1,\ell+2}(\CX)$ with fixed domain and target space independent of $\sigma$. This will enable us to prove the invertibility of $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$  for all sufficiently small  $\gamma>0$ in a fixed small neighborhood of $\sigma=0$ in the closed upper half plane.

Let $s>2, -\frac 32<\ell<-\frac 12$ and $\mathfrak{b}>1$. Then there exists $C_0>0$ such that for all $\gamma>0$ and $\sigma\in\BC, \IM\sigma\geq 0$ with $\gamma+\abs{\sigma}<C_0$, the operator $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma):\mathcal{X}^{s,\ell}(\sigma)\to\eHb^{s-1,\ell+2}(\CX)$ is invertible.

\[
\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}:\eHb^{s-1,\ell+2}(\CX)\to\eHb^{s-1,\ell+2}(\CX),
\]

then it suffices to prove the injectivity of $	\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ provided that $\mathfrak{b}>1$ and $\gamma>0, \IM\sigma\geq 0$ with $\gamma+\abs{\sigma}<C_0$ where $C_0$ is a small constant. To this end, we split the domains and the target spaces as follows

\begin{align*}
\mbox{domain: }\eHb^{s-1,\ell+2}(\CX)&=\mathcal{K}^\perp\oplus\mathcal{K},\quad \mathcal{K}=\ker	\widehat{\mathcal{P}_{0}}(0)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\cap\eHb^{s-1,\ell+2}(\CX)=\langle\tilde{\omega}\rangle\quad\!\!\!\mbox{with}\!\!\quad\! \tilde{\omega}=\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega_{s_0}\\
\mbox{target: }\eHb^{s-1,\ell+2}(\CX)&=\mathcal{R}\oplus\mathcal{R}^\perp,\quad \mathcal{R}=\Ran_{\mathcal{X}^{s,\ell}(0)}\widehat{\mathcal{P}_{0}}(0)
\end{align*}

where $\mathcal{K}^\perp$ and $\mathcal{R}^\perp=\langle\eta\rangle$ are the complementary subspaces to $\mathcal{K}$ and $\mathcal{R}$ respectively inside $\eHb^{s-1,\ell+2}(\CX)$, and $\omega_{s_0}$ is defined as in (<ref>). We notice that

\[
\tilde{\omega}=\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega_{s_0}=\bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{P}_{0}}(0)\bigr)\omega_{s_0}=\widehat{F_gG_{g}\delta_g}\omega_{s_0}\in C^\infty_c(X^\circ)
\]

and we can identify $\mathcal{R}^\perp\cong\BC$ via $c\eta\mapsto\angles{c\eta}{\omega_{s_0}^*}$ where $\omega_{s_0}^*$ is defined as in (<ref>). Under these splittings, $	\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ takes the form

\begin{equation}
\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}=\begin{pmatrix}
\mathcal{W}_{00}(\gamma, \sigma)&	\mathcal{W}_{01}(\gamma, \sigma)\\
\mathcal{W}_{10}(\gamma, \sigma)&	\mathcal{W}_{11}(\gamma, \sigma)
\end{pmatrix}.
\end{equation}

We first list several basic facts: $\mathcal{W}_{01}(0,0), \mathcal{W}_{10}(0,0), \mathcal{W}_{11}(0,0)$ are $0$, and $\mathcal{W}_{00}(0,0)$ is invertible. Since $	\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\in\mathcal{L}(\eHb^{s-1,\ell+2}(\CX),\eHb^{s-1,\ell+2}(\CX))$ is Fredholm of index $0$ (see Lemma <ref>) for $\gamma+\abs{\sigma}$ small and the three components $(0,1),(1,0),(1,1)$ are compact operators (because they all have finite rank), we can conclude that $\mathcal{W}_{00}(\gamma, \sigma)$ is Fredholm of index $0$ as well for $\gamma+\abs{\sigma}$ small.

Suppose $\omega=(\omega_0, \omega_1)\in(\mathcal{K}^\perp\oplus\mathcal{K})\cap\ker\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$, we have $\mathcal{W}_{00}(\gamma, \sigma)\omega_0+\mathcal{W}_{01}(\gamma,\sigma)\omega_1=0$. Assume for the moment that $\mathcal{W}_{00}(\gamma, \sigma)$ is invertible for $\gamma+\abs{\sigma}$ small, then

\begin{equation}\label{eq:roughdis1}
\omega_0=-\mathcal{W}_{00}(\gamma,\sigma)^{-1}\mathcal{W}_{01}(\gamma,\sigma)\omega_1
\end{equation}

and thus

\begin{equation}\label{eq:roughdis2}
\Bigl(\mathcal{W}_{11}(\gamma,\sigma)-\mathcal{W}_{10}(\gamma, \sigma)\mathcal{W}_{00}(\gamma,\sigma)^{-1}\mathcal{W}_{01}(\gamma,\sigma)\Bigr)\omega_1=0.	\end{equation}

We want to show that $\omega_1=0$ and thus $\omega_0=0$ for $\gamma+\abs{\sigma}$.  small. Formally speaking, since $\mathcal{W}_{01}(0,0), \mathcal{W}_{10}(0,0)$ are $0$,  $\mathcal{W}_{01}(\gamma,\sigma), \mathcal{W}_{10}(\gamma, \sigma)$ are of size $\mathcal{O}(\gamma+\abs{\sigma})$. Therefore, it suffices to show that $\mathcal{W}_{11}(\gamma,\sigma)$ is nonzero modulo $\mathcal{O}(\gamma^2+\abs{\sigma}^2)$ for $\gamma+\abs{\sigma}$ small. To make these arguments rigorous, we need to prove the injectivity of $\mathcal{W}_{00}(\gamma,\sigma)$ (and thus the invertibility in view of its $0$ index property) for $\gamma+\abs{\sigma}$ small and the differentiability of $\mathcal{W}_{01}, \mathcal{W}_{10}$ at $(\gamma, \sigma)=(0,0)$, and calculate the Taylor expansion of $\mathcal{W}_{11}(\gamma, \sigma)$ at $(0,0)$ up to the first order.

* Invertibility of $\mathcal{W}_{00}$. We shall prove that for $\gamma+\abs{\sigma}$ small, there exists a uniform constant $C>0$ such that

\begin{equation}\label{eq:uniformnound}
\norm{\mathcal{W}_{00}(\gamma, \sigma)^{-1}}_{\mathcal{R}\to\mathcal{K}^\perp}\leq C.
\end{equation}

The proof is similar to that in the first step of the proof of Lemma <ref>. Specifically, we define $\Pi:\eHb^{s-1,\ell+2}(\CX)\to\mathcal{R}$ to be the projection onto $\mathcal{R}$. Suppose (<ref>) is not true, one can find a sequence $(\gamma_j, \delta_j)\to(0,0)$ and a sequence $\{f_j\}\subset\mathcal{K}^\perp$ with $\norm{f_j}_{\eHb^{s-1,\ell+2}(\CX)}=1$ such that for $\omega_j=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f_j$

\[
1=\norm{f_j}_{\eHb^{s-1,\ell+2}(\CX)}\geq j\norm{\Pi	\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j}_{\eHb^{s-1,\ell+2}(\CX)}
\]

which implies $\Pi	\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j\to 0$ in $\eHb^{s-1,\ell+2}(\CX)$. In view of Theorem <ref>, we have the following uniform Fredholm estimates for $\gamma+\abs{\sigma}$ small

\begin{equation}
\begin{split}
\norm{\omega_j}_{\eHb^{s,\ell}(\CX)}&\leq \tilde{C}\Bigl(\norm{\Pi	\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j}_{\eHb^{s-1,\ell+2}(\CX)}+\norm{(I-\Pi)	\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j}_{\eHb^{s-1,\ell+2}(\CX)}\\
\end{split}
\end{equation}

Using the identification $\mathcal{R}^\perp\cong \BC$ introduced previously, we calculate

\begin{equation}
\begin{split}
&\norm{(I-\Pi)	\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j}_{\eHb^{s-1,\ell+2}(\CX)}\sim\abs{\angles{	\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j}{\omega_{s_0}^*}}=\abs{\angles{	\omega_j}{\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)^*\omega_{s_0}^*}}\\
&\qquad=\abs{\angles{	\omega_j}{\Bigl(\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)^*-\widehat{\mathcal{P}_{0}}(0)^*\Bigr)\omega_{s_0}^*}}\to 0\quad\mbox{as}\quad (\gamma_j, \sigma_j)\to(0,0)
\end{split}
\end{equation}

where we use $\omega_{s_0}^*\in\sHb^{-\frac 12-,\infty}(\CX)$ in the first equality and $\widehat{\mathcal{P}_{0,\gamma_j}}(\sigma_j)^*-\widehat{\mathcal{P}_{0}}(0)^*\to 0$ as bounded operators $\mathcal{L}(\sHb^{1-s,-\ell-2}(\CX),\sHb^{-s,-\ell-1}(\CX))$ in the last step. Since $\omega_j=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f_j$, we have $\norm{\omega_j}_{\eHb^{s,\ell}(\CX)}\sim 1$ and thus $\liminf_{j\to\infty}\norm{\omega_j}_{\eHb^{s_0,\ell_0}(\CX)}\geq \tilde{C}^{-1}\liminf_{j\to\infty}\norm{\omega_j}_{\eHb^{s,\ell}(\CX)}>0$. As a result, there exists a subsequence (not shown in the notation) $\omega_j\to\omega\neq 0$ weakly in $\eHb^{s,\ell}(\CX)$. Then $\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j\to\widehat{\mathcal{W}_{0}}(0)\omega=\widehat{\mathcal{P}_{0}}(0)\omega$ weakly in $\eHb^{s-2,\ell}(\CX)$. Since $\widehat{\mathcal{W}_{0,\gamma_j}}(\sigma_j)\omega_j\to 0$ strongly in $\eHb^{s-1,\ell+2}(\CX)$, we obtain $\widehat{\mathcal{P}_{0}}(0)\omega=0$.

Meanwhile, since $\norm{f_j}_{\eHb^{s-1,\ell+2}(\CX)}=1$, passing to a subsequence, we see that $f_j\to f$ weakly in $\eHb^{s-1,\ell+2}(\CX)$ and strongly in $\eHb^{s-1-\epsilon, \ell+2-\epsilon}(\CX)$. Owing to Lemma <ref> (the continuity in weak operator topology), $\omega_j=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma_j)^{-1}f_j\to \widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}f$ weakly in $\eHb^{s-\epsilon, \ell-\epsilon}$. Therefore, we obtain $0\neq \omega=\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}f$, i.e., $0\neq f=\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega\in\mathcal{K}$. It is clear that one can find $g\in\sHb^{1-s,\ell-2}(\CX)$ such that $g(\mathcal{K}^\perp)=0$ while $g(\tilde{w})=1$ ($\mathcal{K}=\langle \tilde{\omega}\rangle$), but this contradicts the fact that $f_j\to f$ weakly, and thus proves (<ref>). From (<ref>), we conclude that for $\gamma+\abs{\sigma}$ small,  $\mathcal{W}_{00}$ is injective and then the invertibility follows from the fact that $\mathcal{W}_{00}$ has index $0$.

* Differentiability of $\mathcal{W}_{01}, \mathcal{W}_{10}$. We define

\begin{equation}
\mathcal{W}_1(\gamma,\sigma):=\mathcal{W}_{01}(\gamma, \sigma)\oplus\mathcal{W}_{11}(\gamma, \sigma):\mathcal{K}=\langle \tilde{\omega}\rangle\to\eHb^{s-1,\ell+2}(\CX).
\end{equation}

We will show that $\mathcal{W}_1(\gamma, \sigma)$ is continuous for $\gamma+\abs{\sigma}$ small and differentiable at $(0,0)$.

We write

\begin{equation}\label{eq:splitofW1}
\begin{split}
\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}&=\Big(\widehat{\mathcal{W}_{0,\gamma}}(\sigma)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)+\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Big)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\\
\end{split}
\end{equation}

Then it suffices to analyze the first term on the right hand side of (<ref>). For the continuity, since $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\tilde{\omega}\in\eHb^{s,\ell}(\CX)$ depending on $\sigma$ continuously for any $s>2, -\frac 32<\ell<-\frac 12$ (because $\tilde{\omega}\in C^\infty_c(X^\circ)$ and $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ is continuous in $\sigma$ with value in $\mathcal{L}_{\mathrm{op}}(\eHb^{s-1+\epsilon, \ell+2+\epsilon}(\CX),\eHb^{s-\epsilon, \ell-\epsilon}(\CX))$) and  $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\in\mbox{Diff}^1_b$ with $C^\infty_c(X)$ coefficients and depends on $\gamma, \sigma$ smoothly, the continuity of $	\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\tilde{\omega}\in C^\infty_c(X)$ in $(\gamma, \sigma)$ follows

Now we turn to the analysis of the differentiability at $(0,0)$. Since $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)$ are polynomials in $\gamma, \sigma$ with coefficients being first order differential operators, it is certainly differentiable at $(0,0)$. So it suffices to prove that $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\tilde{\omega}$ is differentiable at $\sigma=0$. Now we claim that if $\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\omega\in\rho C^\infty(\CX)+\eHb^{\infty, 1/2-}(\CX)$, then the following resolvent identity holds

\begin{equation}\label{eq:resolventidentity}
\Bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}-\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\Bigr)\omega=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\Bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Bigr)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\omega.
\end{equation}

The proof of the above resolvent identity closely follows <cit.> and uses a regularization argument. Concretely, let $\chi_\epsilon(\rho)=\chi(\epsilon^{-1}\rho)$ where $\chi$ is a nonnegative smooth cutoff such that $\chi=1$ on $[1,\infty)$ and $\chi=0$ on $(-\infty, 1/2]$. With the limit being strong operator limits, we have

\begin{align*}	&\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}-\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\\
&\quad=\lim_{\epsilon\to 0}\Big(\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\chi_\epsilon(\rho)\widehat{\mathcal{W}_{0,\gamma_1}}(0)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\chi_{\epsilon}(\rho)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\Big)\\
&\quad=\lim_{\epsilon\to 0}\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\Big([\chi_\epsilon(\rho),\widehat{\mathcal{W}_{0,\gamma_1}}(0)]+\widehat{\mathcal{W}_{0,\gamma_1}}(0)\chi_{\epsilon}(\rho)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\chi_{\epsilon}(\rho)\Big)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\\
&\quad=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\lim_{\epsilon\to 0}\bigg(\Big(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Big)\chi_{\epsilon}(\rho)\bigg)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\\
&\quad\quad+\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\lim_{\epsilon\to 0}\Big([\chi_\epsilon(\rho),\widehat{\mathcal{W}_{0,\gamma_1}}(0)]\Big)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}:=\Romanupper{1}+\Romanupper{2}.
\end{align*}

As for $\Romanupper{1}$, since $\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\omega\in\rho C^\infty(\CX)+\eHb^{\infty, 1/2-}(\CX)$ which is annihilated modulo $\eHb^{\infty, 3/2-}(\CX)$ by the normal operator $-i\sigma\rho(\rho \pa_\rho-1)$ of $\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)$, it follows that

\begin{align*}
&\quad=\chi_{\epsilon}(\rho)\Big(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Big)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\omega+\epsilon^{-1}\chi'(\epsilon^{-1}\rho)\Big(\rho^3 C^\infty(\CX)+\eHb^{\infty, 5/2-}(\CX)\Big)\\
&\quad\in \chi_{\epsilon}(\rho)\eHb^{\infty, 1/2-}(\CX)+\epsilon^{-1}\chi'(\epsilon^{-1}\rho)\Big(\rho^3 C^\infty(\CX)+\eHb^{\infty, 5/2-}(\CX)\Big).
\end{align*}

Since the $\eHb^{\infty, 1/2-}(\CX)$ norm of the second term $\epsilon\chi'(\epsilon^{-1}\rho)\Big(\rho C^\infty(\CX)+\eHb^{\infty, 1/2-}(\CX)\Big)$ goes to $0$ as $\epsilon\to0$, we arrive at

\[
\Romanupper{1}=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\Big(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Big)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\omega.
\]

For $\Romanupper{2}$, since $\widehat{\mathcal{W}_{0,\gamma_1}}(0)\in\rho^2\mbox{Diff}_b^2$, it follows that $[\chi_\epsilon(\rho), \widehat{\mathcal{W}_{0,\gamma_1}}(0)]$ is uniform (in $\epsilon$) in $\rho^2\mathrm{Diff}_\bop^1$ and $[\chi_\epsilon(\rho), \widehat{\mathcal{W}_{0,\gamma_1}}(0)]$ goes to $0$ in the strong operator topology $\mathcal{L}(\eHb^{s,\ell}(\CX),\eHb^{s-1,\ell+2}(\CX))$, and thus $\Romanupper{2}=0$. This proves (<ref>).

Since $\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\tilde{\omega}=\omega_{s_0}=r^{-1}(dv-dr)\in\rho C^\infty(\CX)+\eHb^{\infty, 1/2-}(\CX)$, applying (<ref>) yields

\begin{equation}
\Bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}-\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\Bigr)\tilde{\omega}=\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\Bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Bigr)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\tilde{\omega}.
\end{equation}

\begin{equation}
\tilde{\omega}(\sigma):=\Bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(0)-\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)\Bigr)\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\tilde{\omega}\in\eHb^{\infty,3/2-}(\CX),
\end{equation}

and write

\begin{equation}
\tilde{\omega}(\sigma)=-\sigma(\pa_{\sigma}\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega_{s_0})+\mathcal{O}_{\eHb^{\infty,3/2-}(\CX)}(\abs{\sigma}^2)
\end{equation}

where the notation $\mathcal{O}_{\eHb^{\infty,3/2-}(\CX)}(\abs{\sigma}^2)$ means that the term has norm in $\eHb^{\infty, 3/2-}(\CX)$ of size $\mathcal{O}(\abs{\sigma}^2)$. Using the fact that $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ is continuous in $\sigma$ with value in $\mathcal{L}_{\mathrm{op}}(\eHb^{s-1+\epsilon, \ell+2+\epsilon}(\CX),\eHb^{s-\epsilon, \ell-\epsilon}(\CX))$, it follows that

\begin{equation}
\Bigl(\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}-\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\Bigr)\tilde{\omega}=-\sigma\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}\pa_{\sigma}\widehat{\mathcal{W}_{0,\gamma_1}}(0)\omega_{s_0}+o_{\eHb^{s,\ell}(\CX)}(\abs{\sigma})
\end{equation}

for any $s>2, -\frac 32<\ell<-\frac 12$. This proves the differentiability of $\mathcal{W}_1$ and thus $\mathcal{W}_{01}$ at $(\gamma, \sigma)=(0,0)$.

As for $\mathcal{W}_{10}$, we proceed in a similar manner. We define

\begin{equation}
\begin{split}
\mathcal{W}_2(\gamma,\sigma):=\mathcal{W}_{10}(\gamma, \sigma)\oplus\mathcal{W}_{11}(\gamma, \sigma):\eHb^{s-1,\ell+2}(\CX)&\to \mathcal{R}^\perp\cong\BC\\
\omega&\mapsto\angles{\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\omega}{\omega^*_{s_0}}
\end{split}
\end{equation}

We will show that $\mathcal{W}_2(\gamma, \sigma)$ is continuous for $\gamma+\abs{\sigma}$ small and differentiable at $(0,0)$.

As $\omega_{s_0}^*=\delta(r-2\Bm_0)dr\in\sHb^{-1/2-,\infty}(\CX)$, we can rewrite

\begin{equation}
\begin{split}
\angles{\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\omega}{\omega^*_{s_0}}&=	\angles{\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\omega}{\widehat{\mathcal{W}_{0,\gamma}}(\sigma)^*\omega^*_{s_0}}\\
\end{split}
\end{equation}

Since $\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ is continuous in $\sigma$ with value in $\mathcal{L}_{\mathrm{op}}(\eHb^{s-1, \ell+2}(\CX),\eHb^{s-\epsilon, \ell-\epsilon}(\CX))$, i.e.,

\[
\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}=
\widehat{\mathcal{W}_{0,\gamma_1}}(0)^{-1}+o_{\mathcal{L}(\eHb^{s-1,\ell+2}(\CX), \eHb^{s-\epsilon,\ell-\epsilon}(\CX))}(1),
\]

and $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)^*-\widehat{\mathcal{W}_{0}}(0)^*$ is a polynomial in $\gamma, \overline{\sigma}$ with coefficients in $\mathcal{L}(\sHb^{1-s,-\ell-2}(\CX),\sHb^{-s,-\ell-1}(\CX))$

\begin{align*}
\end{align*}

it follows that

\begin{equation}\label{eq:diffofW_10}
\begin{split}
\end{split}
\end{equation}

This proves the differentiability of $\mathcal{W}_{10}$ at $(0,0)$.

* Taylor expansion of $\mathcal{W}_{11}$. Since $\mathcal{W}_{11}:\mathcal{K}=\langle\tilde{\omega}\rangle\to\mathcal{R}^\perp\cong\BC$ is given by $\angles{\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}\tilde{\omega}}{\omega^*_{s_0}}$, by (<ref>) we find

\begin{equation}
\mathcal{W}_{11}(\gamma,\sigma)=\sigma\angles{\pa_\sigma\widehat{\mathcal{W}_{0,\gamma}}(\sigma)|_{(0,0)}\omega_{s_0}}{\omega_{s_0}^*}+\gamma\angles{\pa_\gamma\widehat{\mathcal{W}_{0,\gamma}}(\sigma)|_{(0,0)}\omega_{s_0}}{\omega_{s_0}^*}+o(\gamma+\abs{\sigma}).