\end{equation}

Since $\pa_\gamma\widehat{\mathcal{W}_{0,\gamma}}(\sigma)|_{(0,0)}=\gamma^{-1}\bigl(\widehat{\mathcal{W}_{0,\gamma}}(0)-\widehat{\mathcal{P}_{0}}(0)\bigr)$, using the calculation of $\mathcal{W}_{11}^\flat$ in (<ref>) gives

\begin{equation}
\angles{\pa_\gamma\widehat{\mathcal{W}_{0,\gamma}}(\sigma)|_{(0,0)}\omega_{s_0}}{\omega_{s_0}^*}=4\pi(\mathfrak{b}-1).
\end{equation}

We notice that $\pa_\sigma\widehat{\mathcal{W}_{0,\gamma}}(\sigma)|_{(0,0)}=\pa_\sigma\widehat{\mathcal{P}_{0}}(\sigma)|_{\sigma=0}=-i[\mathcal{P}_0, t_{0,*}]$, so we have

\begin{equation}
\begin{split}
\angles{\pa_\sigma\widehat{\mathcal{P}_{0,\gamma}}(\sigma)|_{(0,0)}\omega_{s_0}}{\omega_{s_0}^*}&=-i\angles{[\mathcal{P}_0, t_{0,*}]\omega_{s_0} }{\omega_{s_0}^*}=-i\angles{[\mathcal{P}_0, v]\omega_{s_0} }{\omega_{s_0}^*}=-2\pi i
\end{split}
\end{equation}

where we use

\begin{align*}
[\mathcal{P}_0, v]\omega_{s_0}&=-(G_g\delta_g^*\omega_{s_0})_{\al\be}(\nabla^\al v)+\delta_gG_g(\nabla_\al v\otimes_s\omega_{s_0})\\
\end{align*}

Therefore, we arrive at

\begin{equation}\label{eq:valueofW_11}
\mathcal{W}_{11}(\gamma,\sigma)=-2\pi i\sigma+4\pi (\mathfrak{b}-1)\gamma+o(\gamma+\abs{\sigma}).
\end{equation}

* Final step.
Owing to the previous discussion starting at (<ref>), in order to prove that $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ is injective, it suffices to prove the injectivity of the following operator

\begin{equation}
\mathcal{W}_{11}(\gamma,\sigma)-\mathcal{W}_{10}(\gamma, \sigma)\mathcal{W}_{00}(\gamma,\sigma)^{-1}\mathcal{W}_{01}(\gamma,\sigma):\mathcal{K}\to\mathcal{R}^\perp.
\end{equation}

Combining the differentiability of $\mathcal{W}_{01}, \mathcal{W}_{10}$ at $(0,0)$ and (<ref>) together yields

\[
\mathcal{W}_{11}(\gamma,\sigma)-\mathcal{W}_{10}(\gamma, \sigma)\mathcal{W}_{00}(\gamma,\sigma)^{-1}\mathcal{W}_{01}(\gamma,\sigma)=-2\pi i\sigma+4\pi (\mathfrak{b}-1)\gamma+o(\gamma+\abs{\sigma})
\]

which is nonzero for $\gamma>0, \IM\sigma\geq 0$ with $\gamma+\abs{\sigma}$ small if we set $\mathfrak{b}>1$, and thus is injective. Finally, the invertibility follows from the fact that $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)\widehat{\mathcal{W}_{0,\gamma_1}}(\sigma)^{-1}$ has index $0$ for $\gamma+\abs{\sigma}$ small.

Now we are at the position to prove Proposition <ref>.

The proof proceed in a similar fashion to that in the last step in the proof of Theorem <ref>. Specifically, by Lemma <ref>, there exists $\gamma_0'>0, C'>0$ such that for $0<\gamma<\gamma_0'$, $\IM\sigma\geq 0,\abs{\sigma}<C'$ and $s>2,\ell\in (-\frac 32, \frac 12)$,

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

is invertible. By Proposition <ref>, this implies for $0<\gamma<\gamma_0'$, $\IM\sigma\geq 0,\abs{\sigma}<C'$ and $s>2,\ell<-\frac12, s+\ell>-\frac 12$,

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

is injective as well, and the invertibility follows from the fact that $\widehat{\mathcal{P}_{0,\gamma}}(\sigma)$ has index $0$ (see Lemma <ref>).

Next, according to the high energy estimates (Theorem <ref>) and energy estimate (Proposition <ref> and Theorem <ref>), there exists $\gamma_0''>0, C''>0$ such that for $0<\gamma<\gamma_0''$ and $\IM\sigma\geq 0,\abs{\sigma}>C''$, $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ defined as in (<ref>) is invertible.

For $\sigma$ with $\IM\sigma\geq 0, C'\leq\abs{\sigma}\leq C''$, the fact that $\widehat{\mathcal{W}_{0,0}}(\sigma)=\widehat{\mathcal{P}_{0}}(\sigma)$ (see Proposition <ref>) is invertible and a perturbation argument (as in the final step in the proof of Theorem <ref>) give the invertibility of $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ for $(\gamma, \sigma)$ in an open neighborhood of $(0, \sigma)$. Then the compactness of the region $C'\leq\abs{\sigma}\leq C''$ implies that there exists an $\gamma_0'''>0$ (depending on $C', C''$) such that $\widehat{\mathcal{W}_{0,\gamma}}(\sigma)$ in (<ref>) is invertible for $\IM\sigma\geq 0, C'\leq\abs{\sigma}\leq C''$ and $0<\gamma<\gamma_0'''$. Therefore, the proposition holds for $0<\gamma<\tilde{\gamma}_0$ where $\tilde{\gamma}_0=\min\{\gamma_0',\gamma_0'',\gamma_0'''\}$.

Finally, having Propositions <ref> and <ref> at our disposal, we are able to establish the following theorem for $\widehat{\mathcal{W}_{b,\gamma}}(\sigma)$.

Let $b_0=(\Bm_0, 0,\BQ_0)$ and $\mathfrak{b}>2$. Fix $\gamma\in(0,\tilde{\gamma}_0)$, 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 for

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

\begin{equation}\label{eq:Wmodenonzero}
\widehat{\mathcal{W}_{b_0,\gamma}}(\sigma):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{W}_{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:Wmodezero}
\widehat{\mathcal{W}_{b_0,\gamma}}(0):\{\omega\in\eHb^{s, \ell}(\CX;\scform): 	\widehat{\mathcal{W}_{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 Kerr Newman metric $g_b$ with $b=(\Bm, \Ba,\BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$.

The proof is completely analogous to that of Theorem <ref>.

§.§ Growing  zero modes of the gauge potential wave operator

As discussed around (<ref>) and (<ref>), the zero mode solutions to $\mathcal{W}_{b,\gamma}\omega=0$ on extendible and supported $b$-Sobolev spaces are related to the pure gauge zero mode solutions of $L_{g_b,\gamma}(\dg, \dA)=0$ and dual pure gauge zero mode solutions of $L_{g_b, \gamma}^*(\dg, \dA)=0$, respectively. In this section we will discuss the zero mode solutions to $\mathcal{W}_{b,\gamma}\omega=0$ which are allowed to grow at infinity and are asymptotic to translations and rotations. We will see in the next section that these growing zero modes may generate (dual) pure gauge zero modes solutions to $L^{(*)}_{g_b, \gamma}(\dg,\dA)=0$, which have the expected decay rate at infinity (more precisely, lie in the space $\eHb^{\infty, \ell}(\CX)$ or $\sHb^{-\infty, \ell}(\CX)$ with $-\frac32<\ell<-\frac12$). Throughout this subsection, we focus on the RN metric and slowly rotating metric with sufficiently small charge. Namely, for the parameter $b_0=(\Bm_0, 0, \BQ_0)$ of the RN metric, we assume that
\[\BQ_0\ll\Bm_0.
\]

We first discuss the growing zero modes which are asymptotic to translations.

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

\begin{align}\label{eq:g1form3/2}
\ker \widehat{\mathcal{W}_{b,\gamma}}(0)\cap\eHb^{\infty,-3/2-}(\CX)&=\langle\omega_{b,s_0}\rangle\oplus\{ \omega_{b,s_1}(\sfS):\sfS\in\BFS_1\}, \quad \omega_{b,s_0}=\pa_t^\flat,\\\label{eq:g1formdual3/2}
\ker \widehat{\mathcal{W}_{b,\gamma}}(0)\cap\sHb^{-\infty,-3/2-}(\CX)&=\langle\omega^*_{b,s_0}\rangle\oplus\{ \omega^*_{b,s_1}(\sfS):\sfS\in\BFS_1\},
\end{align}

where $\flat$ denotes the musical isomorphism $V^\flat:=g_b(V,\bullet)$ and $\omega_{b,s_1}(\sfS)$ and $\omega_{b,s_1}^*(\sfS)$ depend on $\sfS\in\BFS_1$ linearly. They satisfy

\begin{equation}\label{eq:symgra3/2}
\delta^*_{g_b}\omega_{b,s_1}\in\eHb^{\infty,1/2-}(\CX;\scsym),\quad G_{g_b}\delta_{g_b}^*\omega^*_{b,\bullet}\in\sHb^{-\infty,1/2-}(\CX;\scsym)
\end{equation}

Moreover, the maps $b\mapsto \omega_{b,s_0}, \omega_{b,s_1}(\sfS)$ and $b\mapsto \omega^*_{b,s_0},\omega^*_{b,s_1}(\sfS)$ can be chosen to be continuous in $b$ with values in the respective spaces.

For later use, we further determine the leading term of $\omega_{b_0,s_1}(\sfS)$ and $\omega^*_{b_0,s_1}(\sfS)$

\begin{equation}\label{eq:refinedg1formmode}
\omega_{b_0,s_1}(\sfS)=du_{b_0,s_1}(\sfS)+\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\gamma+\abs{\BQ_0}^2),\quad \omega^*_{b_0,s_1}(\sfS)=du^*_{b_0,s_1}(\sfS)+\mathcal{O}_{\sHb^{-\infty,-1/2-}(\CX)}(\gamma+\abs{\BQ_0}^2)
\end{equation}

where $u_{b_0,s_1}(\sfS)=(r-\Bm_0)\sfS$ and  $u_{b_0,s_1}^*(\sfS)=(r-\Bm_0)H(r-\ehRN)\sfS$ are defined as in (<ref>).

The proof exploits a normal operator argument as in Propositions <ref> and <ref>. In particular, the normal operator of $-2\widehat{\mathcal{W}_{b,\gamma}}(0)$ is equal to $\widehat{\Box_{\underline{g},1}}(0)$ which is the Euclidean Laplacian $\Delta_{\BR^3}=\rho^2(\rho\pa_\rho(\rho\pa_\rho-1)+\sL)$ tensored with $4\times 4$ identity matrix when working in the standard coordinate trivialization of $\scform$ and annihilates the following $1$-forms

\begin{equation}
dt,\ dx^1,\ dx^2,\ dx^3.
\end{equation}

We note that $dt$ is of scalar type $l=0$ while $dx^i=\sfS dr+rd\sfS$ with $\sfS=x^i/r$ ($i=1,2,3$)  is of scalar type $l=1$.

We first prove (<ref>). Let $u\in\ker\widehat{\mathcal{W}_{b, \gamma}}(0)\cap \sHb^{-\infty,-3/2-}(\CX)$. By the normal operator argument as in Proposition <ref> ( where we shift the contour of integration through the pole $0$ with the space of resonant states given by $\BFS_0$), $\omega$ must take the form $\omega=\chi \omega_0+\tilde{\omega}$ where $\omega_0$ is in the $4$-dimensional space span$\{dt, dx^1, dx^2,dx^3\}$, $\tilde{\omega}\in\eHb^{\infty, -1/2-}(\CX)$ and $\chi$ is a smooth cutoff with $\chi=0$ for $r\leq 3\Bm_0$, $\chi=1$ with $r\geq 4\Bm_0$. This proves that the space in (<ref>) is at most $4$-dimensional. Now we prove that it is indeed $4$-dimensional. We compute for $\omega_0\in\{dt, dx^1, dx^2,dx^3\}$

\[
-\widehat{\mathcal{W}_{b, \gamma}}(0)\chi\omega_0=-(\widehat{\mathcal{W}_{b,\gamma}}(0)-\Box_{\underline{g},1})\chi\omega_0-[\Box_{\underline{g},1}, \chi]\omega_0\in\eHb^{\infty, 3/2-}(\CX)+\eHb^{\infty,\infty}(\CX)=\eHb^{\infty, 3/2-}(\CX).
\]

In view of Theorem <ref>, there exists a unique $\tilde{\omega}\in\eHb^{\infty, -1/2-}(\CX)$ satisfying $\widehat{\mathcal{W}_{b, \gamma}}(0)\tilde{\omega}=-\widehat{\mathcal{W}_{b, \gamma}}(0)\chi \omega_0$. Therefore, $\chi\omega_0+\tilde{\omega}$ gives rise to an element in the space (<ref>) which we denote by $\omega_{b,s_0}$ if $\omega_0=dt$ and $\omega_{b,s_1}(\sfS)$ if $\omega_0=d(r\sfS)\in\{dx^1, dx^2,dx^3\}$. The explicit expression for $\omega_{b,s_0}$ follows from the direction calculation $\widehat{\mathcal{W}_{b,\gamma}}\pa_t^\flat=\delta_{g_b,\gamma}G_g\delta_g^*\pa_t^\flat=0$ since $\pa_t $ is killing (i.e. $\delta_g^*\pa_t^\flat=0$). As for the symmetric gradient, we calculate

\[
\delta_{g_b}^*\omega_{b,s_1}=\delta_{g_b}\tilde{\omega}+(\delta_{g_b}^*-\delta_{\underline{g}}^*)\chi\omega_0+[\delta^*_{\underline{g}},\chi]\omega_0\in\eHb^{\infty,1/2-}(\CX)+\eHb^{\infty,\infty}(\CX)=\eHb^{\infty,1/2-}(\CX)
\]

where we use $\delta_{g_b}^*-\delta_{\underline{g}}^*\in\rho^2\mbox{Diff}_\bop^1, \delta_{g_b}^*\in\rho\mbox{Diff}_\bop^1$ and $\delta_{\underline{g}}^*\omega_0=0$.

Now let us turn to the continuous dependence of $\omega_{b, s_1}(\sfS)$ on $b$.
Suppose $b_j\to b$ and $\omega_{b_j,s_1}(\sfS)=\chi\omega_0+\tilde{\omega}_{b_j}(\sfS)$ where $\omega_0=d(r\sfS)\in\{ dx^1,dx^2,dx^3\}$ and $\tilde{\omega}_{b_j}(\sfS)$ is uniquely determined by $b_j$ and $\sfS$. Let $\omega_{b,s_1}(\sfS)=\chi\omega_0+\tilde{\omega}_b(\sfS)$ and $e_{j,s_1}(\sfS)=\omega_{b_j,s_1}(\sfS)-\omega_{b,s_1}(\sfS)=\tilde{\omega}_{b_j,s_1}(\sfS)-\tilde{\omega}_{b,s_1}(\sfS)$, we find

\[
\widehat{\mathcal{W}_{b, \gamma}}(0)e_{j,s_1}(\sfS)=(\widehat{\mathcal{W}_{b, \gamma}}(0)-\widehat{\mathcal{W}_{b_j, \gamma}}(0))(\tilde{\omega}_{b_j,s_1}(\sfS)+\chi \omega_0).
\]

Since $\{\tilde{\omega}_{b_j,s_1}(\sfS)\}$ is uniformly bounded in $\eHb^{\infty,-1/2-}(\CX)$ and $\widehat{\mathcal{W}_{b, \gamma}}(0)-\widehat{\mathcal{W}_{b_j, \gamma}}(0)\in (b-b_j)\rho^3\mbox{Diff}_b^2$, we see that $\widehat{\mathcal{W}_{b, \gamma}}(0)e_{j,s_1}(\sfS)\in (b-b_j)\eHb^{\infty, 3/2-}$, and thus $e_{j,s_1}(\sfS)\in\eHb^{\infty,-1/2-}(\CX)$ is of size $O(b-b_j)$. This proves the continuous dependence of $\omega_{b,s_1}(\sfS)$ on $b$.

The proof for the dual zero modes $\omega^*_{b,s_0}$ and $\omega^*_{b,s_1}(\sfS$ )proceeds in an analogous manner (we use Theorem <ref> because $\widehat{\mathcal{W}_{b, \gamma}}(0)$ acting on supported $b$-Sobolev space is identical to $\widehat{\mathcal{P}_{b,\gamma}}(0)^*$).

It remains to derive the expressions in (<ref>). Since

\[
\widehat{\mathcal{W}_{b_0,\gamma}}(0)=\delta_{g_{b_0}}G_{g_{b_0}}\delta^*_{g_{b_0}}+F_{g_{b_0}}G_{g_{b_0}}\delta^*_{g_{b_0}}=\frac 12(d\delta_{g_{b_0}}+\delta_{g_{b_0}}d)-\Ric(g_{b_0})+F_{g_{b_0}}G_{g_{b_0}}\delta^*_{g_{b_0}},
\]

we have

\begin{equation}
\begin{split}
\widehat{\mathcal{W}_{b_0,\gamma}}(0)(\omega_{b_0, s_1}(\sfS)-du_{b,s_1}(\sfS))&=\Ric(g)_\al^{\ \be}(du_{b_0,s_1}(\sfS))_\be-F_{g_{b_0}}G_{g_{b_0}}\delta^*_{g_{b_0}}du_{b_0,s_1}(\sfS)\\
\end{split}
\end{equation}

Since $\widehat{\mathcal{W}_{b_0,\gamma}}(0)(\omega_{b_0, s_1}(\sfS)-du_{b,s_1}(\sfS))$ is of scalar type $l=1$ and $\widehat{\mathcal{W}_{b_0,\gamma}}(0)^{-1}$ restricted in scalar type $l=1$ $1$-form spaces exists with norm of size $\mathcal{O}(1)$ (see Remark <ref>), it follows that $\omega_{b_0, s_1}(\sfS)-du_{b,s_1}(\sfS)=\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\gamma+\abs{\BQ_0}^2)$.

Next we turn to the growing zero modes asymptotic to rotations.

For $b=(\Bm, \Ba, \BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$, there exists continuous (in $b$) families

\begin{align}\label{eq:g1form5/2}
b\mapsto \omega_{b,v_1}(\sfV)\in\ker \widehat{\mathcal{W}_{b,\gamma}}(0)\cap\eHb^{\infty,-5/2-}(\CX),\\\label{eq:g1formdual5/2}
b\mapsto \omega^*_{b,v_1}(\sfV)\in\ker \widehat{\mathcal{W}_{b,\gamma}}(0)\cap\sHb^{-\infty,-5/2-}(\CX),
\end{align}

depending linearly on $\sfV\in\BFV_1$, such that

\begin{equation}\label{eq:symmgra5/2}
\delta_{g_b}^*\omega_{b,v_1}(\sfV)\in\eHb^{\infty,1/2-}(\CX;\scsym),\quad 	G_{g_b}\delta_{g_b}^*\omega^*_{b,v_1}(\sfV)\in\sHb^{-\infty,1/2-}(\CX;\scsym).
\end{equation}

More explicitly, we have

\begin{equation}\label{eq:refinedg1formzeromode_rota}
\omega_{b_0,v_1}(\sfV)=r^2\sfV,\quad\omega^*_{b_0,v_1}(\sfV)=r^2H(r-\ehRN)\sfV+\mathcal{O}_{\sHb^{-\infty,-1/2-}(\CX)}(\gamma)
\end{equation}

We first notice that, with respect to RN metric, $r^2\sfV\in\eHb^{\infty,-5/2-}(\CX)$ is dual to the rotation vector field (see the calculation in (<ref>)), and thus killing. Namely, we have $\delta_{g_{b_0}}^*r^2\sfV=0$ and thus $\widehat{\mathcal{W}_{b_0,\gamma}}(0)r^2\sfV=\delta_{g_{b_0,\gamma}}G_{g_{b_0}}\delta^*_{g_{b_0}}r^2\sfV=0$. As for KN metric $g_b$ with $b=(\Bm,\Ba,\BQ)$ near $b_0=(\Bm_0,0,\BQ_0)$, a direct calculation implies

\[
-\widehat{\mathcal{W}_{b, \gamma}}(0)\chi r^2\sfV=-\Bigl(\widehat{\mathcal{W}_{b,\gamma}}(0)-\widehat{\mathcal{W}_{b',\gamma}}(0)\Bigr)\chi r^2\sfV-[\widehat{\mathcal{W}_{b',\gamma}}(0), \chi]r^2\sfV\in\eHb^{\infty, 3/2-}(\CX)+\eHb^{\infty, \infty}(\CX)=\eHb^{\infty, 3/2-}(\CX).
\]

where $b'=(\Bm,0,\BQ)$ and we use $\widehat{\mathcal{W}_{b,\gamma}}(0)-\widehat{\mathcal{W}_{b',\gamma}}(0)\in\rho^4\mbox{Diff}_\bop^2$. In view of Theorem <ref>, there exists a unique $\tilde{\omega}_b(\sfV)\in\eHb^{\infty, -1/2-}(\CX)$ satisfying $\widehat{\mathcal{W}_{b, \gamma}}(0)\tilde{\omega}_b(\sfV)=-\widehat{\mathcal{W}_{b, \gamma}}(0)\chi r^2\sfV$. We then define $\omega_{b,v_1}(\sfV):=\chi r^2\sfV+\tilde{\omega}_b(\sfV)$. For $b=(\Bm,\Ba,\BQ)$, the symmetric gradient $\delta_{g_b}\omega_{b,v_1}(\sfV)$ is given by

\[
\delta_{g_b}^*\omega_{b,v_1}(\sfV)=\delta_{g_b}^*\tilde{\omega}_b(\sfV)+(\delta_{g_b}^*\!\!-\!\delta_{g_{(\Bm,0,\BQ)}}^*)\chi r^2\sfV+[\delta^*_{g_{(\Bm,0,\BQ)}},\chi]r^2\sfV\in\eHb^{\infty,1/2-}(\CX)+\eHb^{\infty,\infty}(\CX)=\eHb^{\infty,1/2-}(\CX)
\]

where we use $\delta_{g_b}^*-\delta_{g_{(\Bm,0,\BQ)}}^*\in\rho^3\mbox{Diff}_\bop^1, \delta_{g_b}^*\in\rho\mbox{Diff}_\bop^1$ and $\delta_{g_{(\Bm,0,\BQ)}}^*r^2\sfV=0$.

The proof of the continuous dependence of $\omega_{b, v_1}$ on $b$ proceeds as in Proposition <ref>. Suppose $b_j=(\Bm_j,\Ba_j,\BQ_j)\to b=(\Bm,\Ba,\BQ)$ and let $\omega_{b_j,v_1}(\sfV)=\chi r^2\sfV+\tilde{\omega}_{b_j}(\sfV)$, $\omega_{b, v_1}(\sfV)=\chi r^2\sfV+\tilde{\omega}_b(\sfV)$ and $e_{j}(\sfV)=\omega_{b_j,v_1}(\sfV)-\omega_{b,v_1}(\sfV)=\tilde{\omega}_{b_j}(\sfV)-\tilde{\omega}_{b}(\sfV)$, we find

\begin{align*}
\widehat{\mathcal{W}_{b, \gamma}}(0)e_{j}(\sfV)&=(\widehat{\mathcal{W}_{b, \gamma}}(0)-\widehat{\mathcal{W}_{b_j, \gamma}}(0))(\tilde{\omega}_{b_j}(\sfV)+\chi r^2\sfV)\\
&=\abs{b-b_j}\rho^3\mbox{Diff}_b^2\tilde{\omega}_{b_j}(\sfV)+\Big(\big(\widehat{\mathcal{W}_{b, \gamma}}(0)-\widehat{\mathcal{W}_{b', \gamma}}(0)\big)-\big(\widehat{\mathcal{W}_{b_j, \gamma}}(0)-\widehat{\mathcal{W}_{b'_j, \gamma}}(0)\big)\Big)\chi r^2\sfV\\
&\qquad+\Big(\widehat{[\mathcal{W}_{b', \gamma}}(0)-\widehat{\mathcal{W}_{b'_j, \gamma}}(0),\chi]r^2\sfV\Big)\\
&=\abs{b-b_j}\rho^3\mbox{Diff}_b^2\tilde{\omega}_{b_j}(\sfV)+\abs{b-b_j}\rho^4\mbox{Diff}_b^2\chi r^2\sfV+\abs{b-b_j}\rho^\infty\mbox{Diff}_b^1r^2\sfV\\
\end{align*}

where $b_j'=(\Bm_j,0,\BQ_j)$ and $b'=(\Bm,0,\BQ)$. As a consequence, $e_j(\sfV)\in\eHb^{\infty,-1/2-}(\CX)$ is of size $O(b-b_j)$. This proves the continuous dependence of $\omega_{b,v_1}(\sfV)$ on $b$.

The statements for dual zero modes $\omega^*_{b,v_1}(\sfV)$ can be proved in a similar way (we use Theorem <ref> because $\widehat{\mathcal{W}_{b, \gamma}}(0)$ acting on supported $b$-Sobolev space is identical to $\widehat{\mathcal{P}_{b,\gamma}}(0)^*$). Here, we present the detail of the derivation of the expression for $\omega^*_{b_0,v_1}(\sfV)$ in (<ref>). First, we calculate

\[
\delta_{g_{b_0}}G_{g_{b_0}}\delta^*_{g_{b_0}} r^2H(r-\ehRN)\sfV=\delta_{g_{b_0}}\Big((r^2\delta(r-\ehRN)dr)\otimes_s\sfV\Big)=-\frac{1}{2r^2}\pa_r(r^4\delta(r-\ehRN)\weight)\sfV=0.
\]

It follows that

\[
\begin{split}
\widehat{\mathcal{W}_{b_0,\gamma}}(0)(\omega^*_{b_0, v_1}(\sfV)-r^2H(r-\ehRN)\sfV)&=-F_{g_{b_0}}\Big((r^2\delta(r-\ehRN)dr)\otimes_s\sfV\Big)=\mathcal{O}_{\eHb^{-\infty,\infty}(\CX)}(\gamma).
\end{split}
\]

Since $\widehat{\mathcal{W}_{b_0,\gamma}}(0)(\omega_{b_0, v_1}(\sfV)-r^2H(r-\ehRN)\sfV)$ is of vector type $l=1$ and $\widehat{\mathcal{W}_{b_0,\gamma}}(0)^{-1}$ restricted in vector type $l=1$ $1$-form spaces exists with norm of size $\mathcal{O}(1)$ (see Remark <ref>), we conclude that
\[\widehat{\mathcal{W}_{b_0,\gamma}}(0)(\omega_{b_0, v_1}(\sfV)-r^2H(r-\ehRN)\sfV)=\mathcal{O}_{\eHb^{-\infty,-1/2-}(\CX)}(\gamma).
\]

§.§ Generalized  zero modes of the gauge potential wave operator

Apart from the growing modes, we also need to take the generalized zero modes  into consideration (the zero modes which allow for polynomial growth in $t_{b,*}$). Before the discussion about the generalized zero modes, we establish the following technical lemma about another family of growing zero modes.

For $b=(\Bm, \Ba, \BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$, there exists continuous (in $b$) families

\begin{align}
b\mapsto \omega^{(1)}_{b,s_1}(\sfS)\in\ker \widehat{\mathcal{W}_{b,\gamma}}(0)\cap\eHb^{\infty,-5/2-}(\CX),\\
b\mapsto \omega^{(1)*}_{b,s_1}(\sfS)\in\ker \widehat{\mathcal{W}_{b,\gamma}}(0)\cap\sHb^{-\infty,-5/2-}(\CX),
\end{align}

depending linearly on $\sfS\in\BFS_1$, such that

\begin{equation}\label{eq:symgra5/2}
\begin{split}
\delta_{g_b}^*\omega^{(1)}_{b,s_1}(\sfS)&=\chi dt\otimes_s d(r\sfS)+\eHb^{\infty,-1/2-}(\CX;\scsym)\\
\delta_{g_b}^*\omega^{(1)*}_{b,s_1}(\sfS)&=\chi dt\otimes_s d(r\sfS)+\sHb^{\infty,-1/2-}(\CX;\scsym)
\end{split}
\end{equation}

Moreover, the leading terms of $\omega^{(1)}_{b,s_1}(\sfS), \omega^{(1)*}_{b,s_1}(\sfS)$are given by

\begin{equation}
\omega_{b,s_1}^{(1)}(\sfS)-\chi r\sfS dt\in\eHb^{\infty,-3/2-}(\CX),\quad \omega_{b,s_1}^{(1)*}(\sfS)-\chi r\sfS dt\in\sHb^{-\infty,-3/2-}(\CX)
\end{equation}

where $\chi$ is a smooth cutoff satisfying $\chi=1$ for $r\geq 4\Bm_0$ and $\chi=0$ for $r\leq 3\Bm_0$.

For later use, we further determine the leading term of $\omega^{(1)}_{b_0,s_1}(\sfS)$

\begin{equation}\label{eq:refinedg1formmode5/2}
\omega^{(1)}_{b_0,s_1}(\sfS)=r\sfS(\weight dt_0-dr)+\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\gamma+\abs{\BQ_0}^2).
\end{equation}

The proof uses the fact that the normal operator of $-2\widehat{\mathcal{W}_{b,\gamma}}(0)$ is equal to $\widehat{\Box_{\underline{g},1}}(0)$, which is the Euclidean Laplacian $\Delta_{\BR^3}=\rho^2(\rho\pa_\rho(\rho\pa_\rho-1)+\sL)$ tensored with $4\times 4$ identity matrix when working in the standard coordinate trivialization of $\scform$ and annihilates the following $1$-form $x^idt=r\sfS dt$ ($i=1,2.3$) where $\sfS=x^i/r\in\BFS_1$. Concretely, we have

\[
-\widehat{\mathcal{W}_{b, \gamma}}(0)\chi r\sfS dt=-\Bigl(\widehat{\mathcal{W}_{b,\gamma}}(0)-\Box_{\underline{g},1}\Bigr)\chi r\sfS dt-[\Box_{\underline{g},1}, \chi]r\sfS dt\in\eHb^{\infty, 1/2-}(\CX)+\eHb^{\infty, \infty}(\CX)=\eHb^{\infty, 1/2-}(\CX).
\]

Since $\ker\widehat{\mathcal{W}_{b, \gamma}}(0)^*\cap\sHb^{-\infty,-1/2+}(\CX)=\{0\}$ by Theorem <ref>, the above equation is solvable. More precisely, there exists a unique $\tilde{\omega}^{(1)}_b(\sfS)\in\eHb^{\infty, -3/2-}(\CX)$ which is in $\ann \{f_1,\cdots, f_4\}$, where $\{f_1,\cdots,f_4\}\subset C^\infty_c(X^\circ)$ is a set of linearly independent functionals on $\widehat{\mathcal{W}_{b_0, \gamma}}(0)\cap\eHb^{\infty,-3/2-}(\CX)$, such that $\widehat{\mathcal{W}_{b, \gamma}}(0)\tilde{\omega}^{(1)}_b(\sfS)=-\widehat{\mathcal{W}_{b, \gamma}}(0)\chi r\sfS dt$. We then define $\omega^{(1)}_{b,s_1}(\sfS):=\chi r\sfS dt+\tilde{\omega}^{(1)}_b(\sfS)$ . As for the symmetric gradient $\delta_{g_b}\omega^{(1)}_{b,s_1}(\sfS)$, we find

\begin{align*}
\delta_{g_b}^*\omega^{(1)}_{b,s_1}(\sfS)&=\chi\delta_{\underline{g}}^*r\sfS dt+[\delta^*_{\underline{g}},\chi]r\sfS dt+(\delta_{g_b}^*\!\!-\!\delta_{\underline{g}}^*)\chi r\sfS dt+\delta_{g_b}^*\tilde{\omega}^{(1)}_b(\sfS)\\
&=\chi dt\otimes_s d(r\sfS)+\eHb^{\infty,\infty}(\CX)+\eHb^{\infty,-1/2-}(\CX)+\eHb^{\infty,-1/2-}(\CX)\\
&=\chi dt\otimes_s d(r\sfS)+\eHb^{\infty,-1/2-}(\CX)
\end{align*}

where we use $\delta_{g_b}^*-\delta_{\underline{g}}^*\in\rho^2\mbox{Diff}_\bop^1$ and $\delta_{g_b}^*\in\rho\mbox{Diff}_\bop^1$.

As for the continuous dependence of $\omega^{(1)}_{b, s_1}(\sfS)$ on $b$, suppose $b_j\to b$ and let $\omega^{(1)}_{b_j,s_1}(\sfS)=\chi r\sfS dt+\tilde{\omega}^{(1)}_{b_j}(\sfS)$, $\omega^{(1)}_{b,s_1}(\sfS)=\chi r\sfS dt+\tilde{\omega}^{(1)}_{b}(\sfS)$ and $e_{j}(\sfS)=\omega^{(1)}_{b_j,s_1}(\sfS)-\omega^{(1)}_{b,s_1}(\sfS)=\tilde{\omega}^{(1)}_{b_j}(\sfS)-\tilde{\omega}^{(1)}_{b}(\sfS)$, we find

\begin{align*}
\widehat{\mathcal{W}_{b, \gamma}}(0)e_{j}(\sfS)&=(\widehat{\mathcal{W}_{b, \gamma}}(0)-\widehat{\mathcal{W}_{b_j, \gamma}}(0))(\tilde{\omega}^{(1)}_{b_j}(\sfS)+\chi r\sfS dt)\in\abs{b-b_j}\eHb^{\infty,1/2-}(\CX).
\end{align*}

As $\widehat{\mathcal{W}_{b,\gamma}}(0)|_{\ann\{f_1,\cdots,f_4\}}:\ann\{f_1,\cdots,f_4\}\to\eHb^{\infty,1/2-}(\CX)$ 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.

The proofs for dual zero modes $\hat{\omega}^{(1)*}_{b,s_1}(\sfS)$ are analogous where we use
\[\ker\widehat{\mathcal{W}_{b, \gamma}}(0)^*\cap\eHb^{\infty, -1/2+}(\CX)=\ker\widehat{\mathcal{P}_{b, \gamma}}(0)\cap\eHb^{\infty, -1/2+}(\CX)=\{0\}.
\]

It remains to derive the expressions in (<ref>). Since

\[
\delta_{g_{b_0}}G_{g_{b_0}}\delta_{g_{b_0}}\big(r\sfS(\weight dt_0-dr)\big)=\frac{\BQ_0^2}{r^3}\sfS(\weight dt_0-dr)\in\BQ_0^2\eHb^{\infty,3/2-}(\CX),
\]

we have

\[
\widehat{\mathcal{W}_{b_0,\gamma}}(0)\Big(\omega^{(1)}_{b_0, s_1}(\sfS)-r\sfS(\weight dt_0-dr)\Big)=\mathcal{O}_{\eHb^{\infty,3/2-}}(\abs{\BQ_0}^2)+\mathcal{O}_{\eHb^{\infty,\infty}(\CX)}(\gamma).
\]

Since the right-hand side of the above equation is of scalar type $l=1$ and $\widehat{\mathcal{W}_{b_0,\gamma}}(0)^{-1}$ restricted in scalar type $l=1$ $1$-form spaces exists with norm of size $\mathcal{O}(1)$ (see Remark <ref>), it follows that $\omega^{(1)}_{b_0, s_1}(\sfS)-r\sfS(\weight dt_0-dr)=\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\gamma+\abs{\BQ_0}^2)$.

Now we are ready to discuss (dual) generalized zero modes which have less decay rate in $r$ at infinity for fixed $t_{b,*}$ and grow linearly in $t_{b,*}$. In the course of the proof, we will find that they are asymptotic to Lorentz boosts.

For $b=(\Bm, \Ba, \BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$, there exists continuous (in $b$) families

\begin{align}\label{eq:ge1formzeromode5/2}
b\mapsto \hat{\omega}_{b,s_1}(\sfS)\in\ker \mathcal{W}_{b,\gamma}\cap\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-5/2-}(\CX),\\\label{eq:ge1formdualzeromode5/2}
b\mapsto \hat{\omega}^*_{b,s_1}(\sfS)\in\ker \mathcal{W}_{b,\gamma}\cap\mbox{Poly}^1(t_{b,*})\sHb^{-\infty,-5/2-}(\CX),
\end{align}

depending linearly on $\sfS\in\BFS_1$, such that

\begin{align}\label{eq:gesymgra}
\delta_{g_b}^*\hat{\omega}_{b,s_1}(\sfS)\in\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-1/2-}(\CX;\scsym),\\\label{eq:gesymgradual}	G_{g_b}\delta_{g_b}^*\hat{\omega}^*_{b,s_1}(\sfS)\in\mbox{Poly}^1(t_{b,*})\sHb^{-\infty,-1/2-}(\CX;\scsym).
\end{align}

Moreover, the leading terms of $\hat{\omega}_{b,s_1}(\sfS)$ are given by

\begin{equation}
\hat{\omega}_{b,s_1}^{(1)}(\sfS)-(tdx^i-x^idt) \in\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-3/2-}(\CX),\quad r\gg 1
\end{equation}

where $\sfS=x^i/r$.

For later use, we further determine the leading term of $\hat{\omega}_{b_0,s_1}(\sfS)$

\begin{equation}\label{eq:refinedge1formmode}
\begin{split}
\hat{\omega}_{b_0,s_1}(\sfS)&=(t_0-r)\omega_{b_0,s_1}(\sfS)-r\sfS(\weight dt_0-dr)+\Bm_0\sfS(dt_0+dr)\\
\end{split}
\end{equation}

We make the ansatz

\begin{equation}
\hat{\omega}_{b,s_1}(\sfS)=t_{\chi_0}\omega_{b,s_1}(\sfS)+\check{\omega}_{b,s_1}(\sfS)
\end{equation}

with $\check{\omega}_{b,s_1}(\sfS)\in\eHb^{\infty,-5/2-}(\CX)$ to be determined. Our aim is to find
\[\hat{\omega}_{b,s_1}\in\ker\mathcal{W}_{b, \gamma}\cap\!\mbox{Poly}^1(t_{b,*})\eHb^{\infty,5/2-}\!(\CX)
\]

such that $\delta_{g_b}^*	\hat{\omega}_{b,s_1}(\sfS)\in\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-1/2-}(\CX)$. To this end, since
\[t_{\chi_0}-t_{b_*}\in\mathcal{A}^{-1}(\CX)\quad \mbox{and} \quad\delta_{g_b}^*\omega_{b,s_1}\in\eHb^{\infty,1/2-}(\CX),
\]

it suffices to solve the following two equations for $\check{\omega}_{b,s_1}(\sfS)\in\eHb^{\infty,-5/2-}(\CX)$

\begin{align}\label{eq:eqnfromsgcondi}
\widehat{\mathcal{W}_{b, \gamma}}(0)\check{\omega}_{b,s_1}(\sfS)=-[\mathcal{W}_{b, \gamma},t_{\chi_0}]\omega_{b,s_1}(\sfS).
\end{align}

We first analyze (<ref>). Since

\[
\]

we further make the ansatz

\begin{equation}\label{eq:defofcheckomega}
\check{\omega}_{b,s_1}(\sfS)=-\omega^{(1)}_{b,s_1}(\sfS)+\tilde{\omega}_{b,s_1}(\sfS)\in\eHb^{\infty,-5/2-}(\CX).
\end{equation}

with $\tilde{\omega}_{b,s_1}(\sfS)\in\eHb^{\infty,-3/2-}(\CX)$ to be determined. Therefore, using $\omega_{b,s_1}(\sfS)=\chi d(r\sfS)+\eHb^{\infty,-1/2-}(\CX)$ and (<ref>), it follows that

\[
\]

and the requirement (<ref>) is satisfied with the choice of $\check{\omega}_{b,s_1}(\sfS)$ defined as in (<ref>).

Next we solve (<ref>) to determine $\tilde{\omega}_{b,s_1}(\sfS)$ in (<ref>). We rewrite (<ref>) as

\begin{equation}\label{eq:eqnfortildeomega}
\begin{split}
&	\widehat{\mathcal{W}_{b, \gamma}}(0)\tilde{\omega}_{b,s_1}(\sfS)=\widehat{\mathcal{W}_{b, \gamma}}(0)\omega^{(1)}_{b,s_1}(\sfS)-[\mathcal{W}_{b, \gamma},t_{\chi_0}]\omega_{b,s_1}(\sfS)\\
&\quad=	\tilde{\delta}_{g_b,\gamma}G_{g_b}\Bigl(\delta_{g_b}^*\omega^{(1)}_{b,s_1}(\sfS)-[\delta_{g_b}^*, t_{\chi_0}]\omega_{b,s_1}(\sfS)\Bigr)-[\tilde{\delta}_{g_b,\gamma},t_{\chi_0}]G_{g_b}\delta_{g_b}^*\omega_{b,s_1}(\sfS)\in\eHb^{\infty,1/2-}(\CX)
\end{split}
\end{equation}

where we use $ \tilde{\delta}_{g_b,\gamma}\in\rho\mbox{Diff}_\bop^1$, $\delta_{g_b}^*\omega^{(1)}_{b,s_1}(\sfS)-[\delta_{g_b}^*,t_{|chi_0}]\omega_{b,s_1}(\sfS)\in\eHb^{\infty,-1/2-}(\CX)$ and
\[\delta_{g_b}^*\omega_{b,s_1}(\sfS)\in\eHb^{\infty,1/2-}(\CX),\quad [\tilde{\delta}_{g_b,\gamma},t_{\chi_0}]\in\mathcal{A}^0(\CX).
\]
Since $\ker\widehat{\mathcal{W}_{b,\gamma}}(0)^*\cap\sHb^{-\infty,-1/2+}(\CX)=\{0\}$, we can solve the above equation (<ref>) for $\tilde{\omega}_{b,s_1}(\sfS)\in\eHb^{\infty,-3/2-}(\CX)$, which is unique modulo $\ker\widehat{\mathcal{W}_{b,\gamma}}(0)\cap\eHb^{\infty,-3/2-}(\CX)$. We then define
\[\hat{\omega}^{(1)}_{b,s_1}(\sfS):=t_{\chi_0}\omega_{b,s_1}(\sfS)-\omega_{b,s_1}^{(1)}(\sfS)+\tilde{\omega}_{b, s_1}(\sfS)
\]

where $\tilde{\omega}_{b,s_1}(\sfS)$ is in the orthogonal complement $\big(\ker\widehat{\mathcal{W}_{b, \gamma}}(0)\cap\eHb^{\infty,-3/2-}(\CX)\big)^\perp$ and thus uniquely determined by $\sfS$ and $b$. As a result, we have for $r\gg1$

\[
\hat{\omega}_{b,s_1}(\sfS)=t\omega_{b,s_1}(\sfS)-\chi r\sfS dt+\eHb^{\infty,-3/2}(\CX)=t d(r\sfS)-r\sfS dt+\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-3/2-}(\CX).
\]

Finally, the continuity in $b$ follows in the same way as in the proof of Lemma <ref>.

The proof for $\hat{\omega}^*_{b,s_1}(\sfS)$ is completely analogous.

It remains to derive the expressions in (<ref>). For the simplicity of calculation, now we let $\tilde{t}=t_0-r$ and make the ansatz

\[
\hat{\omega}_{b_0,s_1}(\sfS)=\tilde{t}\omega_{b_0,s_1}(\sfS)-\omega^{(1)}_{b_0,s_1}+\tilde{\omega}_{b_0,s_1}(\sfS).
\]

Our aim is to solve the following equation for $\tilde{\omega}_{b_0,s_1}(\sfS)\in\eHb^{-3/2-}(\CX)$

\begin{equation}\label{eq:eqfortildeomega}
\begin{split}
\widehat{\mathcal{W}_{b_0, \gamma}}(0)\tilde{\omega}_{b_0,s_1}(\sfS)&=-[\mathcal{W}_{b_0,\gamma}, \tilde{t}\,]\omega_{b_0,s_1}(\sfS)=\frac 12[\Box_{g_{b_0}}, \tilde{t}\,]\omega_{b_0,s_1}(\sfS)+\mathcal{O}_{\eHb^{\infty,\infty}(\CX)}(\gamma)\\
\end{split}
\end{equation}

where we use $[\Box_{g_{b_0}}, \tilde{t}\,]=\Box_{g_{b_0},0}\tilde{t}+2\nabla^\al\tilde{t}\nabla_\al\in\rho^2\mbox{Diff}_\bop^1$ and (<ref>) in the last step. A direct calculation implies

\begin{align*}
&\widehat{\mathcal{W}_{b_0, \gamma}}(0)\Bm_0(dt_0+dr)=\delta_{g_{b_0}}G_{g_{b_0}}\delta^*_{g_{b_0}}\Bm_0(dt_0+dr)+\mathcal{O}_{\eHb^{\infty,\infty}(\CX)}(\gamma)\\
\end{align*}

\begin{align*}
&\widehat{\mathcal{W}_{b_0, \gamma}}(0)\Big(\tilde{\omega}_{b_0,s_1}(\sfS)-\Bm_0(dt_0+dr)\Big)\\
\end{align*}

Meanwhile, with $u=2\Bm_0(2\Bm_0+(\Bm_0-r)\log(r/\Bm_0))$, we have

\begin{align*}
\widehat{\mathcal{W}_{b_0, \gamma}}(0)du&=\frac 12d\delta_{g_{b_0}}du-\Ric(g_0)_\al^{\ \beta}(du)_\beta+F_{g_{b_0}}G_{g_{b_0}}du\\
\end{align*}

Therefore, we obtain

\[
\widehat{\mathcal{W}_{b_0,\gamma}}(0)\Big(\tilde{\omega}_{b_0, s_1}(\sfS)-\Bm_0(dt_0+dr)-du\Big)=\mathcal{O}_{\eHb^{\infty,3/2-}(\CX)}(\abs{\BQ_0}^2+\gamma).
\]

Since the right-hand side of the above equation is of scalar type $l=1$ and $\widehat{\mathcal{W}_{b_0,\gamma}}(0)^{-1}$ restricted in scalar type $l=1$ $1$-form spaces exists with norm of size $\mathcal{O}(1)$ (see Remark <ref>), it follows that $\tilde{\omega}_{b_0, s_1}(\sfS)-\Bm_0(dt_0+dr)-du=\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\gamma+\abs{\BQ_0}^2)$. Therefore, (<ref>) follows from the combination of this fact and (<ref>).

§ MODES OF THE LINEARIZED GAUGE-FIXED EINSTEIN-MAXWELL SYSTEM

In this section, we will discuss the modes of the linearized gauge-fixed Einstein-Maxwell system

\begin{equation}\label{eq:mlgEM}
L_{(g_b,A_b),\gamma}(\dg,\dA)=(2L^E_{(g_b,A_b),\gamma}(\dg,\dA),\ L^M_{(g_b,A_b),\gamma}(\dg,\dA))=0
\end{equation}

\begin{gather}\label{eq:mlgE}
L^E_{(g_b,A_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:mlgM}
L^M_{(g_b,A_b),\gamma}(\dg,\dA)=D_{(g_b, A_b)}(\delta_gdA)(\dg, \dA)+d\delta_{g_b}\dA=0.
\end{gather}

on the RN metric and $4$-electromagnetic potential, i.e., the case $b=b_0=(\Bm_0,0,\BQ_0), \abs{\BQ_0}\ll\Bm_0$.

For the brevity of notation, we rewrite the Lie derivative of a $1$-form $\omega$ with respect to the vector field $X$ as

\[
\wt{\mathcal{L}}_{\omega}X:=\mathcal{L}_{X}\omega
\]

and put

\begin{equation}\label{eq:simplifiedmlgEM}
L_{b,\gamma}:=L_{(g_b,A_b),\gamma},\quad 	L^E_{b,\gamma}:=L^E_{(g_b,A_b),\gamma},\quad	L^M_{b,\gamma}:=L^M_{(g_b,A_b),\gamma}.
\end{equation}

For the various spaces in this section, for instance, $\eHb^{s,\ell}(\CX;E), \sHb^{s,\ell}(\CX;E),\mathcal{A}(\CX;E)$, etc., where $E\to \CX$ is a vector bundle over $\CX$, we will drop the notation $E$ of the bundle if it it clear from the context.

We shall prove that $L_{b_0,\gamma}$ has no non-zero modes in the closed upper half plane for RN metric and $4$-electromagnetic potential $(g_{b_0},A_{b_0})$ (we note that the same result for slowly rotating KN is nontrivial and will be discussed in detail in next section). Moreover, we will describe the space of zero modes and generalized zero modes (with linearly growth in $t_{b,*}$) of $L_{b,\gamma}$ for both RN case and slowly rotating KN case.

§.§ Modes in the closed upper half plane

Let $(g_b,A_b)$ be the KN metric and $4$-electromagnetic potential with $b=(\Bm,\Ba,\BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$ and the linearized gauge-fixed Einstein-Maxwell operator $L_{b,\gamma}:\eHb^{s,\ell}(\CX;\scsym\oplus\scform)\to\eHb^{s,\ell}(\CX;\scsym\oplus\scform)$ be defined as in (<ref>), (<ref>), (<ref>) and (<ref>). In the definition of $\tilde{\delta}_{g_b,\gamma}^*,\tilde{\delta}_{g_b,\gamma}$(see (<ref>), (<ref>) and (<ref>)), $\mathfrak{c}$ is defined as in (<ref>) and $\gamma\in(0, \min\{\gamma_0, \tilde{\gamma}_0\})$ where $\gamma_0,\tilde{\gamma}_0$ are as in Theorem <ref> and <ref> respectively. We also note that we consider the weakly charged case, namely, $\abs{\BQ_0}\ll\Bm_0$ (more precisely, $\abs{\BQ_0}\leq C(\gamma)$ for some small constant $C(\gamma)$). As in the previous sections, we define the Fourier-transformed operator

\begin{equation}
\widehat{L_{b,\gamma}}(\sigma):=e^{i\sigma t_{b,*}}L_{b,\gamma}e^{-i\sigma t_{b,*}}
\end{equation}

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

Let $(g_{b_0}, A_{b_0})$ be the RN metric and $4$-electromagnetic potential where $b_0=(\Bm_0, 0,\BQ_0)$ with $\abs{\BQ_0}\ll\Bm_0$. Then the Fourier-transformed operator $	\widehat{L_{b_0,\gamma}}(\sigma)$ satisfies the following statements:

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

\begin{equation}\label{eq:mlgEMmodenonzero}
\begin{split}
&\widehat{L_{b_0,\gamma}}(\sigma):\{(\dg,\dA)\in\eHb^{s, \ell}(\CX;\scsym\oplus\scform): 	\widehat{L_{b_0,\gamma}}(\sigma)(\dg,\dA)\in\eHb^{s,\ell+1}(\CX; \scsym\oplus\scform)\}\\
\end{split}
\end{equation}

is invertible for $s>3, \ell<-\frac12$ and $s+\ell>-\frac 12$.

* If $s>3$ and $-\frac 32<\ell<-\frac 12$, then stationary operator

\begin{equation}\label{eq:mlgEMmodezero}
\begin{split}
&\widehat{L_{b_0,\gamma}}(0):\{(\dg,\dA)\in\eHb^{s, \ell}(\CX;\scsym\oplus\scform): 	\widehat{L_{b_0,\gamma}}(0)(\dg,\dA)\in\eHb^{s-1,\ell+2}(\CX; \scsym\oplus\scform)\}\\
\end{split}
\end{equation}

has $8$-dimensional kernel and cokernel.

The second statement also holds for the weakly charged and slowly rotating KN metric and $4$-electromagnetic potential $(g_b, A_b)$ with $b=(\Bm, \Ba,\BQ)$ near $b_0=(\Bm_0, 0,\BQ_0)$. Concretely, we have

\begin{align}\label{eq:mlgEMzeromode}
\mathcal{K}_b:=	\ker \widehat{L_{b,\gamma}}(0)\cap\eHb^{\infty,-1/2-}(\CX)&=\{(\dg_{b,s_0},\dA_{b,s_0})\}\oplus\{ (\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS)):\sfS\in\BFS_1\}\\\notag
&\qquad\oplus\{ (\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV)):\sfV\in\BFV_1\},\\\label{eq:mlgEMdualzeromode}
\mathcal{K}_b^*:=	\ker \widehat{L_{b,\gamma}}(0)^*\cap\sHb^{-\infty,-1/2-}(\CX)&=\{(\dg^*_{b,s_0},\dA^*_{b,s_0})\}\oplus\{ (\dg^*_{b,s_1}(\sfS),\dA^*_{b,s_1}(\sfS)):\sfS\in\BFS_1\}\\\notag
&\qquad\oplus\{ (\dg^*_{b,v_1}(\sfV),\dA^*_{b,v_1}(\sfV)):\sfV\in\BFV_1\},
\end{align}

\begin{align}\label{eq:mlgzeromodes0}
(\dg_{b,s_0},\dA_{b,s_0})&=(\dg_{b}(\dot{\Bm},0,\dot{\BQ})+2\delta_{g_b}\tilde{\omega}_{b,s_0},\  \dA_b(0,0,\dot{\BQ})+\wt{\mathcal{L}}_{A_b}\tilde{\omega}_{b,s_0}^{\sharp}+d\phi_{b,s_0}),\quad\dot{\Bm},\dot{\BQ}\in\BR,\\\label{eq:mlgzeromodes1}
(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))&=(2\delta_{g_b}^*\omega_{b,s_1}(\sfS),\  \wt{\mathcal{L}}_{A_b}(\omega_{b,s_1}(\sfS))^{\sharp}+d\phi_{b,s_1}(\sfS)),\\\label{eq:mlgzeromodev1}
(\dg_{b,v_1}(\sfV),\dA^*_{b,v_1}(\sfV))&=(\dg_{b}(0,\dot{\Ba},0)+2\delta_{g_b}^*\tilde{\omega}_{b,v_1}(\sfV),\  \dA_b(0,\dot{\Ba},0)+\wt{\mathcal{L}}_{A_b}(\tilde{\omega}_{b,v_1}(\sfV))^{\sharp}+d\phi_{b,v_1}(\sfV)),
\end{align}

\begin{align}\label{eq:mlgdualzeromodes0}
(\dg^*_{b,s_0},\dA^*_{b,s_0})&=c_1(2G_{g_b}\delta_{g_b}^*\omega^*_{b,s_0},\  4\wt{\mathcal{L}}_{A_b}(\omega^*_{b,s_0})^{\sharp}+d\phi_{b,s_0}^*)+c_2(0, \delta(r-\ehKN)dr),\quad c_1,c_2\in\BR,\\\label{eq:mlgdualzeromodes1}
(\dg^*_{b,s_1}(\sfS),\dA^*_{b,s_1}(\sfS))&=(2G_{g_b}\delta_{g_b}^*\omega^*_{b,s_1}(\sfS),\  4\wt{\mathcal{L}}_{A_b}(\omega^*_{b,s_1}(\sfS))^{\sharp}+d\phi_{b,s_1}^*(\sfS)),\\\label{eq:mlgdualzeromodev1}
(\dg^*_{b,v_1}(\sfV),\dA^*_{b,v_1}(\sfV))&=(2G_{g_b}\delta_{g_b}^*\omega^*_{b,v_1}(\sfV),\  4\wt{\mathcal{L}}_{A_b}(\omega^*_{b,v_1}(\sfV))^{\sharp}+d\phi_{b,v_1}^*(\sfV))
\end{align}

with $\omega_{b,s_1}(\sfS)$, $\omega^*_{b,s_0}$ and $\omega^*_{b,s_1}(\sfS)$, $\omega^*_{b,v_1}(\sfV)$ defined as in (<ref>), (<ref>) and (<ref>).

Moreover, we have
\[(\dg_{b,s_1}(\sfS),\ \dA_{b,s_1}(\sfS)),\quad(\dg_{b,v_1}(\sfV),\ \dA_{b,v_1}(\sfV))\in\eHb^{\infty,1/2-}(\CX),
\]

and the maps $b\mapsto (\dg_{b, \bullet}(\bullet),\ \dA_{b,\bullet}(\bullet))$ and $b\mapsto (\dg^*_{b, \bullet}(\bullet),\ \dA^*_{b,\bullet}(\bullet))$ can be chosen to be continuous in $b$ with values in the respective spaces.

For later use, we further determine the leading term of $(\dg_{b_0,v_1}(\sfV),\ \dA_{b_0,v_1}(\sfV))$

\begin{equation}\label{eq:refinedhv1}
\begin{split}
\dg_{b_0,v_1}(\sfV)&=\Big((-\frac{4\Bm_0}{r}+\frac{2\BQ_0^2}{r^2})dt_0+\frac{4\Bm_0}{r-(\Bm_0-\sqrt{\Bm_0^2-\BQ_0^2})}dr\Big)\otimes_s\sfV+\mathcal{O}_{\eHb^{\infty,1/2-}(\CX)}(\gamma),\\
\dA_{b_0,v_1}(\sfV)&=\mathcal{O}_{\eHb^{\infty,1/2-}(\CX)}(\abs{\BQ_0}).
\end{split}
\end{equation}

In the course of the proof of the above theorem, we will find that the zero modes in $\mathcal{K}_b$ satisfy both the linearized Einstein-Maxwell system and the linearized generalized wave map gauge and Lorenz gauge conditions.

We first analyze the RN case $b=b_0$ and write $(g, A)=(g_{b_0}, A_{b_0})$. Suppose $\widehat{L_{b_0,\gamma}}(\sigma)(\dg,\dA)=0$. Then by Proposition <ref>, $(\dg,\dA)\in\eHb^{\infty,-1/2-}(\CX)$. In view of (<ref>), applying $\delta_g$ to $L^ M_{b_0,\gamma}e^{-i\sigma t_{b_0,*}}(\dg,\dA)=0$ yields

\[
\delta_g d(\delta_ge^{-i\sigma t_{b_0,*}}\dA)=0.
\]

Since $\dA\in\eHb^{\infty,-1/2-}(\CX;\scform)$,  we obtain that  $\delta_g(e^{-i\sigma t_{b_0,*}}\dA)$ is a mode. More precisely, we have
\[
\delta_g(e^{-i\sigma t_{b_0,*}}\dA)\in\begin{cases} e^{-i\sigma t_{b_0,*}}\eHb^{\infty,-1/2-}(\CX;\BC)&\quad\mbox{if}\quad \sigma\neq0,\\
\eHb^{\infty,1/2-}(\CX;\BC)&\quad\mbox{if}\quad \sigma=0.
\end{cases}
\]

By Theorem <ref>, it follows that

\begin{equation}\label{eq:wavepotemtialeqnM}
\delta_g(e^{-i\sigma t_{b_0,*}}\dA)=0,
\end{equation}

and thus $e^{-i\sigma t_{b_0,*}}(\dg, \dA)$ is a mode solution to the Maxwell part of the linearized Einstein-Maxwell system. Applying $\delta_gG_g$ to $L^ E_{b_0,\gamma}e^{-i\sigma t_{b_0,*}}(\dg,\dA)=0$ and using the linearized second Bianchi identity (see the discussion around (<ref>) and (<ref>)) give

\[
\delta_gG_g\tilde{\delta}_{g,\gamma}^*(\tilde{\delta}_{g,\gamma}G_ge^{-i\sigma t_{b_0,*}}\dg)=\mathcal{P}_{b_0,\gamma}(\tilde{\delta}_{g,\gamma}G_ge^{-i\sigma t_{b_0,*}}\dg)=0.
\]

Again, since $\tilde{\delta}_{g,\gamma}G_ge^{-i\sigma t_{b_0,*}}\dg$ is a mode in the same way as $\delta_g(e^{-i\sigma t_{b_0,*}}\dA)$, it follows from Theorem <ref> that

\begin{equation}\label{eq:wavepotemtialeqnE}
\tilde{\delta}_{g,\gamma}G_ge^{-i\sigma t_{b_0,*}}\dg=0,
\end{equation}

and thus $e^{-i\sigma t_{b_0,*}}(\dg, \dA)$ is a mode solution to the linearized Einstein-Maxwell system.

* The case $\IM\sigma\geq 0, \sigma\neq0$. Then we apply the mode stability result Theorem <ref> and find that

\begin{equation}\label{eq:exfornonzeromodesoltoEM}
e^{-i\sigma t_{b_0,*}}(\dg, \dA)=(2\delta^*_g\omega, \mathcal{L}_{\omega^\sharp}A+d\phi)
\end{equation}

where $(\omega,\phi)=e^{-i\sigma t_{b_0,*}}(\omega_0,\phi_0)\in e^{-i\sigma t_{b_0,*}}\eHb^{\infty,\ell'}(\CX;\scform\oplus\BC)$ with $\ell'<-1/2$. Plugging (<ref>) into the equation (<ref>), we find

\[
\widehat{\mathcal{W}_{b_0,\gamma}}(\sigma)\omega_0=0,
\]

and thus $\omega_0=0$ by Theorem <ref>. We then plug (<ref>) with $\omega=0$ into the equation (<ref>) and obtain

\[
\widehat{\Box_g}(\sigma)\phi_0=0.
\]

Therefore, we have $\phi_0=0$ by Theorem <ref>. This proves the injectivity of $\widehat{L_{b_0,\gamma}}(\sigma)$ for $\IM\sigma\geq 0, \sigma\neq0$, and the surjectivity follows from the fact that $\widehat{L_{b_0,\gamma}}(\sigma)$ is Fredholm of index $0$.

* Scalar type $l\geq2$ or vector type $l\geq2$ zero modes. We apply the mode stability result Theorem <ref> and find that

\begin{equation}\label{eq:exforl2zeromodesoltoEM}
(\dg, \dA)=(2\delta^*_g\omega, \mathcal{L}_{\omega^\sharp}A+d\phi)
\end{equation}

where $(\omega,\phi)\in \eHb^{\infty,-3/2-}(\CX;\scform\oplus\BC)$. Plugging (<ref>) into the equation (<ref>), we find

\[
\widehat{\mathcal{W}_{b_0,\gamma}}(0)\omega=0.
\]

Since $\omega\in\eHb^{\infty, -3/2-}(\CX)$ is of scalar or vector type $l\geq 2$, it follows that $\omega=0$ by Theorem <ref> and Proposition <ref>. We then plug (<ref>) with $\omega=0$ into the equation (<ref>) and obtain

\[
\widehat{\Box_g}(0)\phi=0.
\]

Again, since $\phi\in\eHb^{\infty,-3/2-}(\CX)$ with vector or scalar type $l\geq2$, we conclude that  $\phi=0$ by Theorem <ref> and Proposition <ref>. This proves the injectivity of $\widehat{L_{b_0,\gamma}}(0)$ when restricted on scalar or vector type $l\geq 2$ modes.

* Scalar type $l=1$ zero modes. As in the previous case (scalar or vector $l\geq 2$ case), we find that $(\dg, \dA)=(2\delta^*_g\omega, \mathcal{L}_{\omega^\sharp}A+d\phi)$
where $(\omega,\phi)\in \eHb^{\infty,-3/2-}(\CX;\scform\oplus\BC)$ and satisfies first

\[
\widehat{\mathcal{W}_{b_0,\gamma}}(0)\omega=0.
\]

Since $\omega\in\eHb^{\infty, -3/2-}(\CX)$  is of scalar type $l=1$, in view of Proposition <ref> we see that

\begin{equation}\label{eq:exforscalarl1gpE}
\omega=\omega_{b_0,s_1}(\sfS)\quad\mbox{for some}\quad \sfS\in\BFS_1
\end{equation}

where $\omega_{b_0,s_1}(\sfS)$ is defined as in (<ref>). We then plug $\dA=\mathcal{L}_{\omega^\sharp}A+d\phi$ and (<ref>) into the equation (<ref>) and obtain

\[
\]

where we use $\delta_g\in\rho\mbox{Diff}_\bop^1$ and $\mathcal{L}_{(\cdot)^\sharp} A\in\rho^2\mbox{Diff}_\bop^1$ (see (<ref>)). Owing to Theorem <ref> and the fact that $\ker \widehat{\Box_g}(0)\cap\eHb^{\infty,-3/2-}(\CX)$ restricted to scalar type $l=1$ spaces is $0$, there exists a unique $\phi\in\eHb^{\infty,-1/2-}(\CX)$, which we denote by $\phi_{b_0,s_1}(\sfS)$, such that the above equation holds. Therefore, the scalar type $l=1$ nullspace of $\widehat{L_{b_0,\gamma}}(0)$ is $3$-dimensional.  We also note that $\phi_{b_0,s_1}(\sfS)$ is of size $\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\abs{\BQ_0})$ and thus $\dA_{b_0,s_1}(\sfS)=\mathcal{O}_{\eHb^{\infty,1/2-}(\CX)}(\abs{\BQ_0})$. In view of (<ref>), we see that $\dg_{b_0,s_1}(\sfS)=2\delta_g^*\omega_{b_0,s_1}(\sfS)\in\eHb^{\infty,1/2-}(\CX;\scsym)$.

* Scalar type $l=0$ zero modes. If $(\dg,\dA)$ is of scalar type $l=0$, by Theorem <ref> we have

\begin{equation}\label{eq:exforl0zeromode}
(\dg,\dA)=(\dg_{b_0}(\dot{\Bm},0,\dot{\BQ})+2\delta^*_{g}\omega,\  \dA_{b_0}(0,0,\dot{\BQ})+\mathcal{L}_{\omega^\sharp}A+d\phi)
\end{equation}

where $(\omega,\phi)\in \eHb^{\infty,-3/2-}(\CX;\scform\oplus\BC)$. Plugging (<ref>) into (<ref>), it follows that

\[
2\widehat{\mathcal{W}_{b_0, \gamma}}(0)\omega=-\tilde{\delta}_{g,\gamma}G_g\dg_{b_0}(\dot{\Bm},0,\dot{\BQ})\in\eHb^{\infty,1/2-}(\CX).
\]

In view of Theorem <ref>, $\ker\widehat{\mathcal{W}_{b_0,\gamma}}(0)^*\cap\sHb^{-\infty,-1/2+}(\CX;\scform)=\{0\}$. Therefore, we can solve the above equation for $\omega\in\eHb^{\infty,-3/2-}(\CX)$ which is of scalar type $l=0$ and unique modulo $\langle\pa_t^\flat\rangle$. However, since $\delta^*_g\pa_t^\flat=0$ and $\mathcal{L}_{\pa_t}A=0$, a multiple of $\pa_t^\flat$ does not change $\dg$ and the equation for $\phi$ below. This implies $\omega$ is unique and we denote it by $\tilde{\omega}_{b_0,s_0}$. Then we plug $\dA=\dA_{b_0}(0,0,\dot{\BQ})+\mathcal{L}_{\omega^\sharp}A+d\phi$ and $\omega=\tilde{\omega}_{b_0,s_0}$ into the equation (<ref>) and obtain

\[
\]

By Theorem <ref>, we have $\ker\widehat{\Box_g}(0)^*\cap\sHb^{-\infty,-1/2+}(\CX;\BC)=\{0\}$. Therefore, we can solve the above equation for $\phi\in\eHb^{\infty,-3/2-}(\CX;\BC)$ which is of scalar type $l=0$ and unique modulo $\langle1\rangle$. Again, adding a constant to $\phi$ make no change to $\dA$, which means that $\phi$ is unique. We then denote this unique $\phi$ by $\phi_{b_0,s_0}$. This proves that the scalar type $l=0$ kernel of $\widehat{L_{b_0,\gamma}}(0)$ is $2$-dimensional.

* Vector type $l=1$ zero modes. If $(\dg,\dA)$ is of vector type $l=1$, by Theorem <ref> we have

\begin{equation}\label{eq:exforvectorl1zeromode}
(\dg,\dA)=(\dg_{b_0}(0,\dot{\Ba},0)+2\delta^*_{g}\omega,\  \dA_{b_0}(0,\dot{\Ba},0)+\mathcal{L}_{\omega^\sharp}A+d\phi)
\end{equation}

where $(\omega,\phi)\in \eHb^{\infty,-3/2-}(\CX;\scform\oplus\BC)$. Plugging (<ref>) into (<ref>), it follows that

\[
2\widehat{\mathcal{W}_{b_0, \gamma}}(0)\omega=-\tilde{\delta}_{g,\gamma}G_g\dg_{b_0}(0,\dot{\Ba},0)\in\eHb^{\infty,3/2-}(\CX).
\]

In view of Theorem <ref>,
$\ker\widehat{\mathcal{W}_{b_0,\gamma}}(0)^*\cap\sHb^{-\infty,-1/2+}(\CX;\scform)=\{0\}$, and thus the above equation is solvable for $\omega\in\eHb^{\infty,-3/2-}(\CX;\scform)$. Meanwhile, since $\omega$ is of vector type $l=1$, by Proposition <ref>,

there exists a unique $\omega\in\eHb^{\infty,-1/2-}(\CX;\scform)$ of vector type $l=1$ satisfying the above equation. We then denote this unique $\omega$ by $\tilde{\omega}_{b_0,v_1}(\sfV)$. Therefore, we have $\dg_{b_0, v}(\sfV)\in\eHb^{\infty,1/2-}(\CX)$.  Plugging $\dA=\dA_{b_0}(0,\dot{\Ba},0)+\mathcal{L}_{\omega^\sharp}A+d\phi$ and $\omega=\tilde{\omega}_{b_0,v_1}$ into the equation (<ref>), we obtain that

\[
\]

By Theorem <ref>, there exists a unique $\phi\in\eHb^{\infty,-1/2-}$ of vector type $l=1$ satisfying the above equation. We then denote this unique $\phi$ by $\phi_{b_0,v_1}(\sfV)$. Therefore, the vector $l=1$ kernel of $\widehat{L_{b_0,\gamma}}(0)$ is $3$-dimensional. We also note that $\phi_{b_0,v_1}$ is of size $\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX;\BC)}(\abs{\BQ_0})$, and thus $\dA_{b_0,v_1}=\mathcal{O}_{\eHb^{\infty,1/2-}(\CX;\BC)}(\abs{\BQ_0})$.

Now we shall derive the expression for $(\dg_{b_0,v_1}(\sfV),\dA_{b_0,v_1}(\sfV))$ in (<ref>). We write

\[
\dg_{b_0,v_1}(\sfV)=\dg^0_{b_0}(0,\dot{\Ba},0)+2\delta_{g}^*\omega^0.
\]

When rescaling to $\abs{\dot{\Ba}}=1$ and letting $(\theta,\varphi)$ be the spherical coordinates adapted to $\dot{\Ba}$, we can rewrite

\[
\dg_{b_0,v_1}(\sfV)=(2(\weight-1)dt_0-2dr)\otimes_s\sfV+2\delta_{g}^*\omega^0\quad\mbox{with}\quad\sfV=\sin^2\theta d\varphi.
\]

Plugging $\dg_{b_0,v_1}(\sfV)$ into (<ref>) yields

\[
2\widehat{\mathcal{W}_{b_0, \gamma}}(0)\omega^0=-\tilde{\delta}_{g,\gamma}G_g\Big((2(\weight-1)dt_0-2dr)\otimes_s\sfV\Big)=-2r^{-1}\sfV+\mathcal{O}_{\eHb^{\infty,\infty}(\CX)}(\gamma).
\]

We also find that

\[
\delta_g^*f(r)\sfV=\Big(1+\frac{2\Bm_0}{r-(\Bm_0-\sqrt{\Bm_0^2-\BQ_0^2})}\Big)dr\otimes_s\sfV,\quad	\delta_gG_g\delta_g^* f(r)\sfV=-r^{-1}\sfV
\]

\[
\]

Then it follows that

\[
2\widehat{\mathcal{W}_{b_0, \gamma}}(0)(\omega^0-f(r)\sfV)=\mathcal{O}_{\eHb^{\infty,\infty}(\CX)}(\gamma),
\]

and thus $\omega^0-f(r)\sfV=\mathcal{O}_{\eHb^{\infty,-1/2-}(\CX)}(\gamma)$ (according to Remark <ref>, $\widehat{\mathcal{W}_{b_0,\gamma}}(0)^{-1}$ restricted in vector type $l=1$ $1$-form spaces exists with norm of size $\mathcal{O}(1)$ ). This implies

\begin{equation}
\begin{split}
\quad \qquad\dg_{b_0,v_1}(\sfV)&\!\!=(2(\weight-1)dt_0-2dr)\otimes_s\sfV+2\delta_g^*f(r)\sfV\!+\!\mathcal{O}_{\eHb^{\infty,1/2-}(\CX)}(\gamma)\\
\end{split}
\end{equation}

Therefore, $\dg_{b_0,v_1}(\sfV)\in\eHb^{\infty,1/2-}(\CX;\scsym)$.

* Dual zero modes for RN and KN case. Combining the discussion around (<ref>), (<ref>) and (<ref>) with the growing dual zero modes established in <ref>, we see that
\[(2G_{g_b}\delta_{g_b}^*\omega_{b,\bullet}^*(\bullet),\  4\wt{\mathcal{L}}_{A_b}(\omega^*_{b,\bullet}(\bullet))^{\sharp}+d\phi_{b,\bullet}^*(\bullet))\in\ker \widehat{L_{b,\gamma}}(0)^*
\]
provided that

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

For $\omega^*_{b,s_\bullet}(\bullet)$, in view of (<ref>), $\delta_{g_b}\in\rho\mbox{Diff}_\bop^1$ and $\mathcal{L}_{(\bullet)^\sharp}A_b\in\rho^2\mbox{Diff}_\bop^1$, the right-hand side of (<ref>) is in $\sHb^{-\infty,3/2-}(\CX)$. For the case $\phi^*_{b,s_0}$, by Theorem <ref>, there exists a unique $\phi^*_{b,s_0}\in\sHb^{-\infty,-1/2-}(\CX)$ such that (<ref>) holds. We note that adding a multiple of $H(r-\ehRN)$ to $\phi^*_{b,s_0}$ still gives rise to $A^*_{b,s_0}\in\sHb^{-\infty,1/2-}(\CX)$. For the case $\phi^*_{b,s_1}(\sfS), \sfS\in\BFS_1$, by Theorem <ref> again, there exists a unique $\phi^*_{b,s_1}(\sfS)\in\sHb^{-\infty,-1/2-}(\CX)$ such that (<ref>) holds. Therefore, with the above choice of $\phi_{b,\bullet}^*(\bullet)$, we have $\dA^*_{b,s_\bullet}(\bullet)\in\sHb^{-\infty,1/2-}(\CX)$. As for $\omega_{b,v_1}(\sfV)$, according to (<ref>), now the right-hand side of (<ref>) is in $\sHb^{-\infty,1/2-}(\CX)$. Since $\ker\widehat{\Box_{g_b}}(0)\cap\eHb^{\infty,-1/2+}(\CX)=\{0\}$ by Theorem <ref>, (<ref>) is solvable for $\phi_{b,v_1}^*(\sfV)$. More precisely,  there exists a unique $\phi^*_{b,v_1}(\sfV)\in\sHb^{-\infty,-3/2-}(\CX)$, which lies in $\ann(v)$ with $v\in C^\infty_c(X)$ and satisfying $\langle v,H(r-r_b)\rangle\neq 0$, such that (<ref>) holds, and thus $\dA^*_{b,v_1}\in\sHb^{-\infty,-1/2-}(\CX;\scform)$. We also note that by (<ref>) and (<ref>), $G_{g_b}\delta_{g_b}^*\omega^*_{b,\bullet}(\bullet)\in\sHb^{-\infty,1/2-}(\CX;\scsym)$. As a consequence,
\[(2G_{g_b}\delta_{g_b}^*\omega_{b,\bullet}^*(\bullet),\  4\wt{\mathcal{L}}_{A_b}(\omega^*_{b,\bullet}(\bullet))^{\sharp}+d\phi_{b,\bullet}^*(\bullet))\quad \mbox{and}\quad(0, \delta(r-\ehKN)dr)
\]

indeed span a $8$-dimension subspace of $\mathcal{K}^*_b$.

* Zero modes for KN case. First, the above construction of the dual zero modes of $\widehat{L_{b,\gamma}}(0)^*$ and the index $0$ property of $\widehat{L_{b,\gamma}}(0)$ imply that $\mathcal{K}_b$ is at least $8$-dimensional. Next, we shall prove by a contradiction argument that it is at most $8$-dimensional for $b$ near $b_0$, and thus must be $8$-dimensional. Suppose $\{h_1,\cdots,h_8\}$ is a basis of $\mathcal{K}_{b_0}$ and choose $h_1^\flat,\cdots, h_8^\flat\in C^\infty_c(X^\circ;\scsym\oplus \scform)$ such that $\angles{h_i}{h_j^\flat}=\delta_{ij}$. (Consider the map $\Phi: C^\infty_c(X^\circ)\to\BC^{8}$ defined by $\Phi(h^\flat)=(\langle h_{1},h^\flat \rangle,\cdots,\langle h_{8},h^\flat\rangle)$, and note that $\Phi$ is surjective. Otherwise, there would exist some $a=(a_1, \cdots,a_{8})\neq 0\in\BC^{8}$ such that $a\cdot\Phi(h^\flat)=\langle\sum_{i=1}^{8}a_i h_{i},h^\flat\rangle=0$ for all $h^\flat\in C^\infty_c(X^\circ)$, which implies $\langle \sum_{i=1}^{8}a_i h_{i}, h^\flat\rangle=0$ all $h^\flat\in\sHb^{-\infty,-\ell}(\CX)$ since $C^\infty_c(X^\circ) $ is dense in $\sHb^{-\infty,-\ell}(\CX)$. However, this is impossible because $\sum_{i=1}^{8}a_i h_{i}\neq 0$). Assume for the sake of contradiction that there exists a sequence $b_j\to 0$ such that for all $j$, the dimension of $\mathcal{K}_b$ is greater than $8$. Then one can find $u_j\in\mathcal{K}_{b_j}$ such that $u_j\in\mbox{ann}\{h^\flat_1,\cdots,h^\flat_8\}$ with $\norm{u_j}_{\eHb^{\infty,-1/2-}(\CX)}=1$. As in the proof of the first step of Lemma <ref> (using the Fredholm estimates for $\widehat{L_{b,\gamma}}(0)$), there exists a subsequence $u_j\to u\neq0$ weakly in $\eHb^{\infty,-1/2-}(\CX)$ and then $u\in\ker\widehat{L_{b_0,\gamma}}(0)$. Therefore, one can pick $h^\flat\in\mbox{span}\{h_1^\flat,\cdots,h_8^\flat\}$ such that $\angles{u}{h^\flat}=1$, but $0=\angles{u_j}{h^\flat}\to\angles{u}{h^\flat}=1$, which leads to a contradiction.

Getting the explicit expressions for $(\dg_{b,\bullet}(\bullet),\dA_{b,\bullet}(\bullet))$ as in (<ref>), (<ref>) and (<ref>) requires a direct argument. First,  by (<ref>), we indeed have $(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))=(2\delta_{g_b}\omega_{b,s_1}(\sfS),\ \wt{\mathcal{L}}_{	A_b}(\omega_{b,s_1}(\sfS))^\sharp+d\phi_{b,s_1}(\sfS) )\in\ker\widehat{L_{b,\gamma}}\cap\eHb^{\infty,1/2-}(\CX)$ for suitable $\phi_{b,s_1}$. It remains to construct $(\dg_{b,s_0},\dA_{b,s_0})$ and $(\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV))$ which extend $(\dg_{b_0,s_0},\dA_{b_0,s_0})$ and $(\dg_{b_0,v_1}(\sfV),\dA_{b_0,v_1}(\sfV))$, respectively. We make the ansatz

\begin{equation}
(\dg_{b,s_0},\dA_{b,s_0})=(\dot{g}_b(\dot{\Bm},0,\dot{\BQ})+2\delta^*_{g_b}\omega,\  \dA_b(0,0,\dot{\BQ})+\mathcal{L}_{\omega^\sharp}A_b+d\phi)
\end{equation}

with $\omega, \phi$ to be found. Since 	$(\dg_{b,s_0},\dA_{b,s_0})$ of the form in the above ansatz satisfies the linearized Einstein-Maxwell system, we have $	(\dg_{b,s_0},\dA_{b,s_0})\in\ker\widehat{L_{b,\gamma}}(0)$ provided that the generalized linearized gauge conditions are satisfied, namely,

\begin{equation}\label{eq:mgaugecondieqn}
\tilde{\delta}_{g_b,\gamma}G_{g_b}\dg_{b,s_0}=0\quad\mbox{and}\quad \delta_{g_b}d\dA_{b,s_0}=0.
\end{equation}

Then as in the proof of scalar type $l=0$ zero modes of $\widehat{L_{b_0,\gamma}}(0)$, we can find a unique $\omega_{b,s_0}\in\eHb^{\infty,-3/2-}(\CX;\scform)$, which lies in $\ann\{f_1,\cdots,f_4\}$ with $\{f_1,\cdots,f_4\}\subset C^\infty_c(X)$ and being a set of linearly independent functionals on $\ker\widehat{\mathcal{W}_{b_0, \gamma}}(0)\cap\eHb^{\infty,-3/2-}(\CX;\scform)$, and $\phi_{b,s_0}\in\eHb^{\infty,-3/2-}(\CX;\BC)$, which lies in $\ann(v)$ with $v\in C^\infty_c(X)$ and satisfying $\langle 1,v\rangle\neq 0$, such that (<ref>) holds. The construction of $(\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV))$ is analogous. We make the ansatz

\begin{equation}
(\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV))=(\dot{g}_b(0,\dot{\Ba},0)+2\delta^*_{g_b}\omega,\  \dA_b(0,\dot{\Ba},0)+\mathcal{L}_{\omega^\sharp}A_b+d\phi)
\end{equation}

where $\sfV\in\BFV_1$ is dual to the rotation around the axis $\dot{\Ba}$ and $\omega,\phi$ are to be found. Then by using the discussion above and proceeding as in the proof of vector type $l=1$ zero modes of $\widehat{L_{b_0,\gamma}}(0)$, the conclusion (<ref>) follows and $(\dg_{b,v_1}(\sfV), \dA_{b,v_1}(\sfV))\in\eHb^{\infty, 1/2-}(\CX)$.

* Continuity. The continuity (in $b$) of the zero modes and dual zero modes follows from the explicit construction above (see the proof of continuity in <ref>) and that of $\omega_{b,s_1}(\sfS)$ and $\omega^*_{b, \bullet}(\bullet)$.

§.§ Generalized zero modes: linear growth

Now we turn to discussing the generalized zero mode solutions of $L_{b,\gamma}(\dg,\dA)=0$. Suppose $(\dg,\dA)=(\sum^k_{j=1}t^j_{b,*}\dg_j,\sum^k_{j=1}t^j_{b,*}\dA_j)$ and $L_{b,\gamma}(\dg,\dA)=0$. Due to the fact that $[L_{b,\gamma},\pa_{t_{b,*}}]=0$, we find that $\pa_{t_{b,*}}^j(\dg,\dA)\in\ker L_{b,\gamma}$ for $0\leq j\leq k$. Taking $j=k$, we obtain that the leading order term $(\dg_k,\dA_k)$ of $(\dg,\dA)$ lies in the $\ker \widehat{L_{b,\gamma}}(0)$. Next we further determine what the leading order term $(\dg_k,\dA_k)$ should be.

Let $k\geq 1$. Then there does not exist  $(\dg,\dA)=(\sum^k_{j=1}t^j_{b,*}\dg_j,\sum^k_{j=1}t^j_{b,*}\dA_j)$, with $0\neq(\dg_k,\dA_k)\in\{(\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV)):\sfV\in\BFV_1\}$ and $(\dg_j,\dA_j)\in\eHb^{\infty,-1/2-}(\CX;\scsym\oplus\scform)$, such that $L_{b,\gamma}(\dg,\dA)=0$.

According to the discussion above, it suffices to consider the case $k=1$. Assume
\[(\dg,\dA)=(t_{b,*}\dg_{b,v_1}(\sfV)+\dg_0,\  t_{b,*}\dA_{b,v_1}(\sfV)+\dA_0)
\]

with $\sfV\neq 0$. Then we need to show that there is no $(\dg_0,\dA_0)\in\eHb^{\infty,-1/2-}(\CX)$ such that the following equation holds

\[
\widehat{L_{b,\gamma}}(0)(\dg_0,\dA_0)=-[L_{b,\gamma},\ t_{b,*}](\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV))\in\eHb^{\infty,3/2-}(\CX)
\]

where we use $[L_{b,\gamma},\ t_{b,*}]\in\rho\mbox{Diff}_\bop^1$ and $(\dg_{b,v_1}(\sfV),\dA_{b,v_1}(\sfV)\in\eHb^{\infty,1/2-}(\CX)$. To this end, we need to verify that the pairing of the right-hand side with $(\dg^*_{b,v_1}(\sfV),\dA_{b,v_1}^*(\sfV))$ is non-zero. We note that it suffices to check the RN case $b=b_0$ because the general KN case with $b$ near $b_0$ follows directly by the continuity (in $b$). Using the expression (<ref>) for $(\dg_{b_0,v_1}(\sfV),\dA_{b_0,v_1}(\sfV))$, we find that

\begin{equation}
\begin{split}
&\quad\angles{[L_{b_0,\gamma},\ t_{b_0,*}](\dg_{b_0,v_1}(\sfV),\dA_{b_0,v_1}(\sfV))}{(\dg^*_{b_0,v_1}(\sfV'),\dA^*_{b_0,v_1}(\sfV'))}\\
% &=\angles{[L_{b_0,\gamma},\ t_{b_0,*}](\dg_{b_0,v_1}(\sfV),0)}{(\dg^*_{b_0,v_1}(\sfV'),\dA^*_{b_0,v_1}(\sfV'))}+\mathcal{O}(\abs{\BQ_0})\\
&=\angles{[L^E_{b_0,\gamma},\ t_{b_0,*}](\dg_{b_0,v_1}(\sfV),0)}{(\dg^*_{b_0,v_1}(\sfV'),0)}+\mathcal{O}(\abs{\BQ_0})\\
&=\angles{[L^E_{b_0,\gamma},\ t_{b_0,*}](\dg_{b_0,v_1}(\sfV),0)}{(2G_g\delta^*_g(r^2H(r-\ehRN)\sfV'),0)}+\mathcal{O}(\abs{\BQ_0}+\gamma)
\end{split}
\end{equation}

where we use
(<ref>) in the last step. Since $G_{g_{b_0}}\delta_{g_{b_0}}(r^2H(r-\ehRN)\sfV')=r^2\delta(r-\ehRN)dr\otimes_s\sfV'$ is supported at $r=\ehRN$ where $t_{b_0,*}=t_0$, we can replace $t_{b_0,*}$ by $t_0$ in the subsequent calculation. We then compute

\begin{equation}
\begin{split}
&\quad	\angles{[L^E_{b_0,\gamma},\ t_0](\dg_{b_0,v_1}(\sfV),0)}{\ 2r^2\delta(r-\ehRN)dr\otimes_s\sfV'}\\
&=\angles{[L^E_{b_0,\gamma},\ t_0](4\Bm_0r^{-1}(dr-dt_0)\otimes_s\sfV,0)}{\ 2r^2\delta(r-\ehRN)dr\otimes_s\sfV'}+\mathcal{O}(\abs{\BQ_0}^2+\gamma)\\
&=-\angles{[\Box_{g_{b_0},2},\ t_0](4\Bm_0r^{-1}(dr-dt_0)\otimes_s\sfV)}{\ 2r^2\delta(r-\ehRN)dr\otimes_s\sfV'}+\mathcal{O}(\abs{\BQ_0}^2+\gamma)\\
&=\angles{8\Bm_0r^{-2}\big((dr-(1+\frac{\Bm_0}{r})dt_0)\big)\otimes_s\sfV}{\ 2r^2\delta(r-\ehRN)dr\otimes_s\sfV'}+\mathcal{O}(\abs{\BQ_0}^2+\gamma)\\
\end{split}
\end{equation}

where we use (<ref>) in the first step, (<ref>) and (<ref>) in the second step and $[\Box_{g_{b_0},2},\ t_0]h=h\Box_{g_{b_0}}t_0+2(\nabla^\al t_0) \nabla_\al h$ in the third step. Therefore, the pairing is indeed non-zero if $\sfV=\sfV'$ and $\abs{\BQ_0}+\gamma$ is sufficiently small.

We now exclude the generalized zero modes whose leading order term is $(\dg_{b,s_0},\dA_{b,s_0})$.

Let $k\geq 1$. Then there does not exist  $(\dg,\dA)=(\sum^k_{j=1}t^j_{b,*}\dg_j,\sum^k_{j=1}t^j_{b,*}\dA_j)$, with $0\neq(\dg_k,\dA_k)=(\dg_{b,s_0},\dA_{b,s_0})$ and $(\dg_j,\dA_j)\in\eHb^{\infty,-1/2-}(\CX;\scsym\oplus\scform)$, such that $L_{b,\gamma}(\dg,\dA)=0$.

As discussed in the proof of Lemma <ref>, it suffices to prove the statement for the case $k=1$. We first consider the case $b=b_0$. Suppose

\[(\dg,\dA)=(t_{b_0,*}\dg_{b_0,s_0}+\dg_0,\  t_{b_0,*}\dA_{b_0,s_0}+\dA_0)\in\ker L_{b_0,\gamma}
\]

with $(\dg_0,\dA_0)\in\eHb^{\infty,-1/2-}(\CX)$. Applying $\delta_{g_{b_0}}$ to $L^M_{b_0,\gamma}(\dg,\dA)=0$ gives $\delta_{g_{b_0}}\dA\in\ker\Box_{g_{b_0}}$. As pointed out in Remark <ref>,  we have $\delta_{g_{b_0}}\dA_{b_0,s_0}=0$. This implies

\[
\ker\widehat{\Box_{g_{b_0}}}(0)\ni\delta_{g_{b_0}}\dA=[\delta_{g_{b_0}},t_{b_0,*}]\dA_{b_0,s_0}+\delta_{g_{b_0}}\dA_0\in\eHb^{\infty,-1/2-}(\CX;\BC).	\]

Therefore, $\delta_{g_{b_0}}\dA=0$ by Theorem <ref>. That is, $(\dg,\dA)$ satisfies the Maxwell part of the linearized Einstein-Maxwell system. Then as in the proof of Theorem <ref>, we apply $\delta_{g_{b_0}}G_{g_{b_0}}$ to $L^E_{b_0,\gamma}(\dg,\dA)=0$ and use the linearized second Bianchi identity to obtain $\tilde{\delta}_{g_{b_0},\gamma}G_{g_{b_0}}\dg\in\ker\mathcal{P}_{b_0,\gamma}$. Since $\tilde{\delta}_{g_{b_0},\gamma}G_{g_{b_0}}\dg_{b_0,s_0}=0$ (see Remark <ref>), it follows that

\[
\ker\widehat{\mathcal{P}_{b_0,\gamma}}(0)\ni\tilde{\delta}_{g_{b_0},\gamma}G_{g_{b_0}}\dg=[\tilde{\delta}_{g_{b_0},\gamma}G_{g_{b_0}},t_{b_0,*}]\dg_{b_0,s_0}+\tilde{\delta}_{g_{b_0},\gamma}G_{g_{b_0}}\dg_0\in\eHb^{\infty,-1/2-}(\CX;\scform),
\]

and thus $\tilde{\delta}_{g_{b_0},\gamma}G_{g_{b_0}}\dg=0$ by Theorem <ref>. As a consequence, $(\dg,\dA)$ satisfies the linearized Einstein-Maxwell system. By the statement (5) in mode stability Theorem <ref>, we have

\[
(\dg,\dA)=(\dg_{b_0}(\dot{\Bm},0,\dot{\BQ})+2\delta^*_{g_{b_0}}\omega,\ \dA_{b_0}(0,0,\dot{\BQ})+\mathcal{L}_{\omega^\sharp}A_{b_0}+d\phi)
\]

where $(\omega,\phi)\in\mbox{Poly}^2(t_{b_0,*})\eHb^{\infty,-3/2-}(\CX;\scform\oplus\BC)$. To make the calculation simpler, we work in $(t_0=t+r_{b_0,*}, r,\theta,\varphi)$ coordinates instead.  Since $\dg$ is linear in $t_0$, it follows that the coefficient of $t_0^2$ in $\omega$ and $\phi$ must be a multiple of $\pa_{t_0}^\flat$ and $1$, respectively. Then equating the coefficient of $t_0$ of the equality $\dg=\dg_{b_0}(\dot{\Bm},0,\dot{\BQ})+2\delta^*_{g_{b_0}}\omega$ yields

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

where $c\in\BR$ and $\tilde{\omega}\in\eHb^{\infty, \ell'}(\CX;\scform)$ for some $\ell'\in\BR$ is of scalar type $l=0$. Assume $\tilde{\omega}=f(r)dt_0+g(r)dr$ and let $S=c(dt_0)\otimes_s\pa_{t_0}^\flat+\delta^*_{g_{b_0}}\tilde{\omega}$. A direct calculation implies $0=S_{rr}=g'(r)$ and thus $g(r)=m$ for some $m\in\BR$. Then $0=S_{t_0r}=\frac c2+\frac 12 f'(r)+\frac m2 \weight'$ implies $f(r)=-cr+\frac{2m\Bm_0}{r}-\frac{m\BQ_0^2}{r^2}+n$ for some $n\in\BR$. Next $0=S_{\theta\theta}=r(f(r)+m\weight)=-cr^2+(m+n)r$ gives $c=0,m+n=0$, and thus $\tilde{\omega}=-m\weight dt_0+mdr$. Finally, $ \frac{2\dot{\Bm}}{r}-\frac{2\BQ_0\dot{\BQ}}{r^2}=S_{t_0t_0}=\frac{m}{2}\weight\weight'-\frac{m}{2}\weight\weight'=0$ implies that $\dot{\Bm}=\BQ_0\dot{\BQ}=0$ and thus the leading term $g_{b_0,s_0}$ of $\dg$ is $0$. If $\dot{\BQ}\neq 0, \BQ_0=0$, equating the coefficient of $t_0$ of the equality $\dA=\dA_{b_0}(0,0,\dot{\BQ})+d\phi$ yields

\begin{equation}
\frac{\dot{\BQ}}{r}dt_0=cdt_0+d\tilde{\phi}
\end{equation}

where $c\in\BR$ and $\tilde{\phi}\in\eHb^{\infty,\ell'}(\CX;\BC)$ for some $\ell'\in\BR$ is of scalar type $l=0$. This implies that $\dot{\BQ}=0$ and thus  the leading term $A_{b_0,s_0}$ of $\dA$ is $0$.
This proves the lemma for the RN case $b=b_0$.

This non-existence result for $b=b_0$ in turn means that the following equation

\[
\widehat{L_{b_0,\gamma}}(0)(\dg_0,\dA_0)=-[L_{b_0,\gamma},\ t_{b_0,*}](\dg_{b_0,s_0},\dA_{b_0,s_0})\in\eHb^{\infty,3/2-}(\CX)
\]

cannot be solved for $(\dg_0,\dA_0)\in\eHb^{\infty,-1/2-}(\CX;\scsym\oplus\scform)$ (The right-hand side is in the stated Sobolev space because by Proposition <ref>, $(\dg_{b_0,s_0},\dA_{b_0,s_0})=\rho C^\infty(\CX)+\eHb^{\infty,1/2-}(\CX)$ whose leading term $\rho C^\infty(\CX)$ is annihilated by the normal operator $2\rho(\rho\pa_\rho-1)$ of $-[L_{b_0,\gamma},\ t_{b_0,*}]$). Suppose $\{(\dg^1_{b, s_0},\dA^1_{b,s_0}),\,(\dg^2_{b, s_0},\dA^2_{b,s_0})\}$ is a basis of $\{(\dg_{b, s_0},\dA_{b,s_0})\}$ and  $\{(\dg^{*1}_{b, s_0},\dA^{*1}_{b,s_0}),\,(\dg^{*2}_{b, s_0},\dA^{*2}_{b,s_0})\}$ is a basis of $\{(\dg^*_{b, s_0},\dA^*_{b,s_0})\}$.  In terms of pairing,  we have

\[
\angles{[L_{b_0,\gamma},\ t_{b_0,*}](\dg^i_{b_0,s_0},\dA^i_{b_0,s_0})}{(\dg^{*j}_{b_0,s_0},\dA^{*j}_{b_0,s_0})},\quad 1\leq i,j\leq 2
\]

is non-degenerate.
Due to the continuity of $(\dg_{b,s_0},\dA_{b,s_0})$ and $(\dg^*_{b,s_0},\dA^*_{b,s_0})$ in $b$, this pairing is still non-zero for KN case with $b$ near $b_0$. This finishes the proof of the lemma.

However, there do exist generalized zero modes (growing linearly in $t_{b,*}$) whose leading order term is given by $(\dg_{b,s_1}(\sfS), \dA_{b,s_1}(\sfS))$ for $\sfS\in\BFS_1$.

Let $b_0=(\Bm_0,0,\BQ_0)$ with $\abs{\BQ_0}\ll\Bm_0$. Then the following space

\begin{align}
\widehat{\mathcal{K}}_{b_0}&:=\ker L_{b_0,\gamma}\cap\mbox{Poly}^1(t_{b_0,*})\eHb^{\infty,-1/2-}(\CX;\scsym\oplus\scform).
\end{align}

is $11$-dimensional. The $3$-dimensional quotient space	$\widehat{\mathcal{K}}_{b_0}/\mathcal{K}_{b_0}$ is spanned by the following continuous (in $b$) families

\begin{equation}
\{(\hat{g}_{b_0,s_1}(\sfS),\hat{A}_{b_0,s_1}(\sfS)):\ \sfS\in\BFS_1\}.
\end{equation}

\begin{equation}
(\hat{g}_{b_0,s_1}(\sfS),\hat{A}_{b_0,s_1}(\sfS))=(2\delta_{g_{b_0}}^*\hat{\omega}_{b_0,s_1}(\sfS),\ \wt{\mathcal{L}}_{A_{b_0}}(\hat{\omega}_{b_0,s_1}(\sfS))^{\sharp}+d\hat{\phi}_{b_0,s_1}(\sfS))
\end{equation}

where $\hat{\omega}_{b_0,s_1}(\sfS)$ is defined as in (<ref>) and $\hat{\phi}_{b_0,s_0}(\sfS)\in\mbox{Poly}^1(t_{b_0,*})\eHb^{\infty,-3/2-}(\CX;\BC)$ depends linearly in $\sfS\in\BFS_1$.

The generalized zero modes $(\hat{g}_{b_0,s_1}(\sfS),\hat{A}_{b_0,s_1}(\sfS))$ extend to continuous families in $b$

\begin{equation}
b\to(\hat{g}_{b,s_1}(\sfS),\hat{A}_{b,s_1}(\sfS))\in \ker L_{b,\gamma}\cap\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-1/2-}(\CX;\scsym\oplus\scform)
\end{equation}

\begin{equation}
(\hat{g}_{b,s_1}(\sfS),\hat{A}_{b,s_1}(\sfS))=(2\delta_{g_{b}}^*\hat{\omega}_{b,s_1}(\sfS),\ \wt{\mathcal{L}}_{A_{b}}(\hat{\omega}_{b,s_1}(\sfS))^{\sharp}+d\hat{\phi}_{b,s_1}(\sfS))
\end{equation}

where $\hat{\omega}_{b,s_1}(\sfS)$ is defined as in (<ref>) and $\hat{\phi}_{b,s_0}(\sfS)\in\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-3/2-}(\CX;\BC)$ depends linearly in $\sfS\in\BFS_1$. Moreover, $(\hat{g}_{b,s_1}(\sfS),\hat{A}_{b,s_1}(\sfS))$ satisfies both the linearized Einstein-Maxwell system and the linearized generalized wave map gauge and generalized Lorenz gauge conditions.

Meanwhile, there exists continuous families (in $b$)

\begin{equation}
b\to(\hat{g}_{b,s_1}^*(\sfS),\hat{A}_{b,s_1}^*(\sfS))\in\ker L_{b,\gamma}^*\cap\mbox{Poly}^1(t_{b,*})\sHb^{-\infty,-1/2-}(\CX;\scsym\oplus\scform)
\end{equation}

\begin{equation}
(\hat{g}_{b,s_1}^*(\sfS),\hat{A}_{b,s_1}^*(\sfS))=(2G_{g_b}\delta_{g_b}^*\hat{\omega}_{b,s_1}^*(\sfS),\ 4\wt{\mathcal{L}}_{A_{b_0}}(\hat{\omega}^*_{b,s_1}(\sfS))^{\sharp}+d\hat{\phi}^*_{b,s_1}(\sfS))
\end{equation}

where $\hat{\omega}_{b,s_1}^*(\sfS)$ is defined as in (<ref>) and $\hat{\phi}^*_{b,s_0}(\sfS)\in\mbox{Poly}^1(t_{b,*})\sHb^{\infty,-3/2-}(\CX;\BC)$.

Let $(\dg,\dA)=(t_{b,*}\dg_1+\dg_0, t_{b,*}\dA_1+\dA_0)\in\ker L_{b,\gamma}$ with $(\dg_{j},\dA_{j})\in\eHb^{\infty,-1/2-}(\CX)$.

* Generalized zero modes in the RN case $b=b_0$.  In view of Lemma <ref> and <ref>, $(\dg_1,\dA_1)$ must take the form $(2\delta_{g_{b_0}}\omega_{b_0,s_1}(\sfS), \wt{\mathcal{L}}_{A_{b_0}}(\omega_{b_0,s_1}(\sfS))^\sharp+d\phi_{b_0,s_1}(\sfS))$ for some $\sfS\in\BFS_1$. By the argument at the beginning of the proof of Lemma  <ref>, we see that $(\dg,\dA)$ satisfies the generalized linearized gauge condition, that is,

\begin{equation}\label{eq:temmlgc}
\tilde{\delta}_{g_{b_0}}G_{g_{b_0}}\dg=0,\quad \delta_{g_{b_0}}\dA=0.
\end{equation}

Therefore, $(\dg,\dA)$ solves the linearized Einstein-Maxwell system. Since

\begin{equation}\label{eq:exforgA}
\begin{split}
\dg&=2\delta^*_{g_{b_0}}(t_{b_0,*}\omega_{b_0,s_1}(\sfS))-2[\delta^*_{g_{b_0}},t_{b_0,*}]\omega_{b_0,s_1}(\sfS)+\dg_0=2\delta^*_{g_{b_0}}(t_{b_0,*}\omega_{b_0,s_1}(\sfS))+\dg_0' , \\
\dA&=\wt{\mathcal{L}}_{A_{b_0}}(t_{b_0,*}\omega_{b_0,s_1}(\sfS))^\sharp+d\big(t_{b_0,*}\phi_{b_0,s_1}(\sfS)\big)-\Big(\iota_{\omega_{b_0,s_1}(\sfS)^\sharp}A_{b_0}+\phi_{b_0,s_1}(\sfS)\Big)dt_{b_0,*}+\dA_0\\
\end{split}
\end{equation}

\begin{align*}
\dg_0'&=\dg_0-2[\delta^*_{g_{b_0}},t_{b_0,*}]\omega_{b_0,s_1}(\sfS)\in\eHb^{\infty,-3/2-}(\CX),\\
\dA_0'&=\dA_0-\Big(\iota_{\omega_{b_0,s_1}(\sfS)^\sharp}A_{b_0}+\phi_{b_0,s_1}(\sfS)\Big)dt_{b_0,*}\in\eHb^{\infty,-1/2-}(\CX).
\end{align*}

where we use $[\delta^*_{g_{b_0}},t_{b_0,*}]\in\mathcal{A}^0(\CX), \omega_{b_0,s_1}(\sfS)\in\eHb^{\infty,-3/2-}(\CX)$ and $\phi_{b_0,s_1}(\sfS)\in\eHb^{\infty,-1/2-}(\CX)$, it follows that $(\dg_0',\dA_0')$ solves the linearized Einstein-Maxwell system as well. By the statement (4) in Theorem <ref>, we have

\begin{equation}\label{eq:exforg'A'}
(\dg_0',\dA_0')=(2\delta_{g_{b_0}}^*\omega,\  \mathcal{L}_{\omega^{\sharp}}A_{b_0}+d\phi)
\end{equation}

with $(\omega,\phi)\in\eHb^{\infty,-5/2-}(\CX;\scform\oplus\BC)$. Plugging (<ref>) and (<ref>) into (<ref>) yields

\begin{align}\label{eq:eqnforEpotential}
\widehat{\mathcal{W}_{b_0, \gamma}}(0)\omega&=-[\mathcal{W}_{b_0, \gamma},\  t_{b_0,*}]\omega_{b_0,s_1}(\sfS)\in\eHb^{\infty,-1/2-}(\CX),\\\label{eq:eqnforMpotential}
&\quad+[\Box_{g_{b_0}},\  t_{b_0,*}]\phi_{b_0,s_1}(\sfS)-[\delta_{g_{b_0}},t_{b_0,*}]\wt{\mathcal{L}}_{A_{b_0}}(\omega_{b_0,s_1}(\sfS))^\sharp
\end{align}

We first consider (<ref>). Since $\ker\widehat{\mathcal{W}_{b_0, \gamma}}(0)^*\cap\sHb^{-\infty,1/2+}(\CX)=\{0\}$ by Theorem <ref>, it follows that (<ref>) can be solved for $\omega\in\eHb^{\infty,-5/2-}(\CX)$. According to Proposition <ref>, we know that

\[
\hat{\omega}_{b_0,s_1}(\sfS)=t_{b_0,*}\omega_{b_0,s_1}(\sfS)+\check{\omega}_{b_0,s_1}(\sfS)\in\ker\mathcal{W}_{b_0, \gamma}
\]

with $\check{\omega}_{b_0,s_1}(\sfS)\in\eHb^{\infty,-5/2-}(\CX)$. Therefore,

\[
\widehat{\mathcal{W}_{b_0, \gamma}}(0)\big(\omega-\check{\omega}_{b_0,s_1}(\sfS)\big)=-\mathcal{W}_{b_0, \gamma}\hat{\omega}_{b_0,s_1}(\sfS)=0
\]

and $\omega-\check{\omega}_{b_0,s_1}(\sfS)\in \mbox{span}\{\omega^{(1)}_{b_0,s_1}(\sfS):\  \sfS\in\BFS_1\}$ (In fact $\ker\widehat{\mathcal{W}_{b_0, \gamma}}(0)\cap\eHb^{\infty,-5/2-}(\CX)$ restricted to the scalar type $l=1$ $1$-form is equal to $\mbox{span}\{\omega_{b_0,s_1}(\sfS),\omega^{(1)}_{b_0,s_1}(\sfS):\  \sfS\in\BFS_1\}$. Since $(\dg,\dA)$ is in the quotient space $\widehat{\mathcal{K}}_{b_0}/\mathcal{K}_{b_0}$, we can exclude $\omega_{b_0,s_1}(\sfS)$). Using (<ref>), we rewrite

\begin{align*}
\dg&=2\delta^*_{g_{b_0}}\hat{\omega}_{b_0,s_1}(\sfS)+2\delta^*_{g_{b_0}}\Big(\omega-\check{\omega}_{b_0,s_1}(\sfS)\Big)\\
\end{align*}

Since we require $\dg\in\mbox{Poly}^1(t_{b_0,*})\eHb^{\infty,-1/2-}(\CX)$, it follows that $\delta^*_{g_{b_0}}\big(\omega-\check{\omega}_{b_0,s_1}(\sfS)\big)$ must lie in $\eHb^{\infty,-1/2-}(\CX)$. Owing to (<ref>), $\omega-\check{\omega}_{b_0,s_1}(\sfS)$ must be $0$. That it, $\dg=2\delta_{g_{b_0}}\hat{\omega}_{b_0,s_1}(\sfS)$.

Next we turn to (<ref>). Since $\omega=\check{\omega}_{b_0,s_1}(\sfS)\in\eHb^{\infty,-5/2}(\CX)$ and $\phi_{b_0,s_1}\in\eHb^{\infty,-1/2-}(\CX)$, the right-hand side of (<ref>) lies in $\eHb^{\infty,1/2-}(\CX)$. By Theorem <ref> ($\ker\widehat{\Box_{g_{b_0}}}(0)^*\cap\sHb^{-\infty,-1/2+}(\CX)=\{0\}$) and Proposition <ref> ($\ker\widehat{\Box_{g_{b_0}}}(0)\cap\eHb^{\infty,-3/2-}(\CX)$ restricted to scalar type $l=1$ is $0$), there exists a unique $\phi\in\eHb^{\infty,-3/2-}(\CX)$ which we denote by $\check{\phi}_{b_0,s_1}(\sfS)$ satisfying (<ref>). This proves that $\widehat{\mathcal{K}}_{b_0}/\mathcal{K}_{b_0}$ is at most $3$-dimensional.

Using $\dg=2\delta_{g_{b_0}}^*\hat{\omega}_{b_0,s_1}(\sfS)\in\mbox{Poly}^1(t_{b_0,*})\eHb^{\infty,-1/2}(\CX)$ (see (<ref>)) and rewriting
\begin{equation}\label{eq:vefiofdecayofA}
\begin{split}
\dA&=\wt{\mathcal{L}}_{A_{b_0}}(\hat{\omega}_{b_0,s_1}(\sfS))^\sharp+d\big(t_{b_0,*}\phi_{b_0,s_1}(\sfS)+\check{\phi}_{b_0,s_1}(\sfS)\big)\\
\end{split}
\end{equation}

we conclude that $(\dg,\dA)=(\hat{g}_{b_0,s_1}(\sfS),\hat{A}_{b_0,s_1}(\sfS))=(2\delta_{g_{b_0}}^*\hat{\omega}_{b_0,s_1}(\sfS), \wt{\mathcal{L}}_{A_{b_0}}(\hat{\omega}_{b_0,s_1}(\sfS))^\sharp+d(\hat{\omega}_{b_0,s_1}(\sfS)))$ with $\hat{\phi}_{b_0,s_1}(\sfS)=t_{b_0,*}\phi_{b_0,s_1}(\sfS)+\check{\omega}_{b_0,s_1}(\sfS)\in\mbox{Poly}^1(t_{b_0,*})\eHb^{\infty,-3/2}(\CX)$ indeed lies in $\widehat{\mathcal{K}}_{b_0}$.

* Generalized zero modes in the KN case. For $b$ near $b_0$, we make the ansatz

\[
(\dg,\dA)=(2\delta_{g_{b}}^*\hat{\omega}_{b,s_1}(\sfS), \wt{\mathcal{L}}_{A_{b}}(\hat{\omega}_{b,s_1}(\sfS))^\sharp+d(t_{b,*}\phi_{b,s_1}(\sfS))+d\phi).
\]

with $\phi\in\eHb^{\infty,-3/2}(\CX)$ to be determined. By the arguments in the proof of RN case above (the last two paragraphs), there exists $\check{\phi}_{b,s_1}(\sfS)\in\eHb^{\infty,-3/2}(\CX)$ such that with $\hat{\phi}_{b,s_1}(\sfS)=t_{b,*}\phi_{b,s_1}(\sfS)+\check{\phi}_{b,s_1}(\sfS)\in\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-3/2}(\CX)$,

\begin{align*}
&=(2\delta_{g_{b}}^*\hat{\omega}_{b,s_1}(\sfS), \wt{\mathcal{L}}_{A_{b}}(\hat{\omega}_{b,s_1}(\sfS))^\sharp+d(\hat{\phi}_{b,s_1}(\sfS)))\in\ker L_{b,\gamma}\cap\mbox{Poly}^1(t_{b,*})\eHb^{\infty,-1/2-}(\CX)
\end{align*}

extends $(\hat{g}_{b_0,s_1}(\sfS), \hat{A}_{b_0,s_1}(\sfS))$ and is continuous in $b$.

* Generalized dual zero modes for RN and KN cases. Combining the discussion around (<ref>), (<ref>) and (<ref>) with the generalized dual zero mode $\hat{\omega}_{b,s_1}^*(\sfS)=t_{b,*}\omega_{b,s_1}^*(\sfS)+\check{\omega}_{b,s_1}^*(\sfS)$ of $\mathcal{W}_{b,\gamma}$ established in Proposition <ref>, we see that
\[(\hat{g}^*_{b,s_1}(\sfS),\hat{A}^*_{b,s_1}(\sfS))=(2G_{g_b}\delta_{g_b}^*\hat{\omega}_{b,s_1}^*(\sfS),\  4\wt{\mathcal{L}}_{A_b}(\hat{\omega}^*_{b,s_1}(\sfS))^{\sharp}+d(t_{b,*}\phi^*_{b,s_1}(\sfS))+d\check{\phi}_{b,s_1}^*(\sfS))\in\ker L_{b,\gamma}^*
\]
provided that

\begin{equation}\label{eq:eqnforgedualpotentialM}
\begin{split}
&\quad+[\Box_{g_{b}},\  t_{b,*}]\phi^*_{b,s_1}(\sfS)-4[\delta_{g_{b}},t_{b,*}]\wt{\mathcal{L}}_{A_{b}}(\omega^*_{b,s_1}(\sfS))^\sharp\in\sHb^{-\infty,1/2-}(\CX).
\end{split}
\end{equation}

In view of Theorem <ref> and Proposition <ref>, $\ker\widehat{\Box_{g_{b}}}(0)\cap\eHb^{\infty,-1/2+}(\CX)=\{0\}$, and thus there exists a unique $\check{\phi}^*_{b,s_1}(\sfS)\in\sHb^{-\infty,-3/2-}(\CX)$ (up to addition of a multiple of $H(r-\ehKN)$)) such that (<ref>) holds. We put $\hat{\phi}^*_{b,s_1}(\sfS)=t_{b,*}\phi^*_{b,s_1}+\check{\phi}^*_{b,s_1}(\sfS)$ and find that $
\hat{A}_{b,s_1}(\sfS)\in\mbox{poly}^1(t_{b,*})\sHb^{-\infty,-1/2-}(\CX)$ by the argument in (<ref>). We also recall that $G_{g_b}\delta_{g_b}^*\hat{\omega}^*_{b,s_1}(\sfS)\in\mbox{Poly}^1(t_{b,*})\sHb^{-\infty,-1/2-}(\CX)$ (see (<ref>)). As a consequence,  the continuous (in $b$) families $(\hat{g}^*_{b,s_1}(\sfS),\hat{A}^*_{b,s_1}(\sfS))$ indeed lie in $\ker L_{b,\gamma}^*\cap\mbox{poly}^1(t_{b,*})\sHb^{-\infty,-1/2-}(\CX)$.

\begin{align}
(\check{g}_{b,s_1}(\sfS),\ \check{A}_{b,s_1}(\sfS))&:=(\hat{g}_{b,s_1}(\sfS)-t_{b,*}\dg_{b,s_1}(\sfS),\  \hat{A}_{b,s_1}(\sfS)-t_{b,*}\dA_{b,s_1}(\sfS))\in\eHb^{\infty,-1/2-}(\CX),\\
(\check{g}^*_{b,s_1}(\sfS),\ \check{A}^*_{b,s_1}(\sfS))&:=(\hat{g}^*_{b,s_1}(\sfS)-t_{b,*}\dg^*_{b,s_1}(\sfS),\  \hat{A}^*_{b,s_1}(\sfS)-t_{b,*}\dA^*_{b,s_1}(\sfS))\in\sHb^{-\infty,-1/2-}(\CX).
\end{align}

be the coefficients of $t_{b,*}^0$ of the generalized (dual) zero modes constructed above. For later use, we determine the leading order behavior of them.

For $\sfS\in\BFS_1$, we have

\begin{equation}\label{eq:leadingtermofgAhat}
\begin{split}
(\check{g}_{b,s_1}(\sfS),\ \check{A}_{b,s_1}(\sfS))&\in\rho C^\infty(\CX)+\eHb^{\infty,1/2-}(\CX),\\ (\check{g}^*_{b,s_1}(\sfS),\ \check{A}^*_{b,s_1}(\sfS))&\in\rho C^\infty(\CX)+\sHb^{-3/2-C(a,\gamma),1/2-}(\CX)
\end{split}
\end{equation}

where $C(a,\gamma)>0$ is a sufficiently small constant depending on $a,\gamma$.

Arguing as in the proof of Proposition <ref>, we again exploit the normal operator argument. First, since $(\dg_{b,s_1}(\sfS),\ \dA_{b,s_1}(\sfS))\in\eHb^{\infty,1/2-}(\CX)$, according to Proposition <ref> (where we shift the contour of integration through the pole $\xi=-2i$ and thus the space of resonant states is $\BFS_1$), it follows that

\[
(\dg_{b,s_1}(\sfS),\ \dA_{b,s_1}(\sfS))\in \rho^2\Omega_1+\eHb^{\infty,3/2-}(\CX)
\]

\[
\Omega_1\in\mbox{span}\{\sfS dt^2,\ \sfS dt\otimes_s dx^i,\ \sfS dx^i\otimes_s dx^j,\ \sfS dt,\ \sfS dx^i\mid \sfS\in\BFS_1,\ 1\leq i,j\leq 3\}.
\]

\[
-\widehat{L_{b,\gamma}}(0)	(\check{g}_{b,s_1}(\sfS),\ \check{A}_{b,s_1}(\sfS))=[L_{b,\gamma}, t_{b,*}](\dg_{b,s_1}(\sfS),\ \dA_{b,s_1}(\sfS))\in-2\rho^3\Omega_1+\eHb^{\infty, 5/2-}(\CX)
\]

where we use the fact $[L_{b,\gamma}, t_{b,*}]\in-2\rho(\rho\pa_\rho-1)+\rho^2\mbox{Diff}_\bop^1$, it follows that we can solve the above equation for $(\check{g}_{b,s_1}(\sfS),\ \check{A}_{b,s_1}(\sfS))$ by first solving away the leading term $-2\rho^3\Omega_1$ and then the error term which lies in $\eHb^{\infty, 5/2-}(\CX)$.

For the leading term $2\rho^3\Omega_1$, proceeding as in the proof of Proposition <ref> (here we have $k=-1, l=1$), we need to solve the following equation

\[
\Big(\rho\pa_\rho(\rho\pa_\rho -1)+\sL\Big)(\dg_1,\dA_1)=-2\rho\Omega_1.
\]

Since now $k(k+1)=(-1)\cdot 0\neq 1\cdot 2=l(l+1)$, we conclude that $(\dg_1, \dA_2)\in(-1/2)\cdot(-2\rho\Omega_1)=\rho\Omega_1$.

For the error term, we need to solve the following equation

\[
\Big(\rho\pa_\rho(\rho\pa_\rho -1)+\sL\Big)(\dg_2,\dA_2)=f\in \eHb^{\infty, 5/2-}(\CX),\quad (\dg_2,\dA_2)\in\eHb^{\infty, -1/2}(\CX).
\]

Arguing as in the proof of Proposition <ref>, we see that $(\dg_2,\dA_2)\in\eHb^{\infty, -1/2}(\CX)\in \rho C^\infty(\pa_+\CX)+\eHb^{\infty, 1/2-}(\CX)$. This proves (<ref>) for $(\check{g}_{b,s_1}(\sfS),\ \check{A}_{b,s_1}(\sfS))$.

For $(\check{g}^*_{b,s_1}(\sfS),\ \check{A}^*_{b,s_1}(\sfS))$, the above proof also applies. As for the regularity of $(\check{g}^*_{b,s_1}(\sfS),\ \check{A}^*_{b,s_1}(\sfS))$ near the radial point at event horizon (by the same reasoning as in the proof of statement (2) of Proposition <ref>, $(\check{g}^*_{b,s_1}(\sfS),\ \check{A}^*_{b,s_1}(\sfS))$ is smooth away from the radial point at even horizon), we notice that $[L_{b,\gamma}, t_{b,*}](\dg^*_{b,s_1}(\sfS),\ \dA^*_{b,s_1}(\sfS))$ is one derivative less regular than $(\dg^*_{b,s_1}(\sfS),\ \dA^*_{b,s_1}(\sfS))$ and $(\widehat{L_{b,\gamma}}(0)^*)^{-1}$ gains one derivative. Therefore, $(\check{g}^*_{b,s_1}(\sfS),\ \check{A}^*_{b,s_1}(\sfS))$ has at least the same regularity as $(\dg^*_{b,s_1}(\sfS),\ \dA^*_{b,s_1}(\sfS))$, which is $-\frac32-C(a,\gamma)$ in view of Proposition <ref>.

§.§ Generalized zero modes: quadratic growth

Finally, we exclude the generalized zero mode solutions with quadratic growth in $t_{b,*}$ of $L_{b,\gamma}$.

Let $k\geq 2$. Then there does not exist  $(\dg,\dA)=(\sum^k_{j=1}t^j_{b,*}\dg_j,\sum^k_{j=1}t^j_{b,*}\dA_j)$, with $0\neq(\dg_k,\dA_k)=\{(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS)): \sfS\in\BFS_1\}$ and $(\dg_j,\dA_j)\in\eHb^{\infty,-1/2-}(\CX;\scsym\oplus\scform)$, such that $L_{b,\gamma}(\dg,\dA)=0$.

According to the discussion at the beginning of the previous subsection, it suffices to consider the case $k=2$. For the simplicity of calculation, we replace $t_{b,*}$ by another smooth function $t_1:=t_0-2r$. Since $t_{b,*}-t_1\in\mathcal{A}^{0-}(\CX)$, we assume
\[(\dg,\dA)=(t_1^2\dg_{b,s_1}(\sfS)+t_1\dg_1+\dg_0,\  t_1^2\dA_{b,s_1}(\sfS)+t_1\dA_1+\dA_0)
\]

with $\sfS\neq 0$ and $(\dg_{j},\dA_{j})\in\eHb^{\infty,-1/2-}(\CX)$ for $j=0,1$. We write $L_{b,\gamma}(\dg,\dA)$ as a polynomial of degree $1$ in $t_1$ as follows

\begin{align*}
L_{b,\gamma}(\dg,\dA)&=t_1\Big(2[L_{b,\gamma},\  t_1](\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))+L_{b,\gamma}(\dg_1,\dA_1)\Big)\\
&\quad+\Big([[L_{b,\gamma},\  t_1],\ t_1](\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))+[L_{b,\gamma},\  t_1](\dg_1,\dA_1)+L_{b,\gamma}(\dg_0,\dA_0)\Big).
\end{align*}

Then $L_{b,\gamma}(\dg,\dA)=0$ is equivalent to the vanishing of the coefficients of the above polynomial. Namely,

\begin{align}\label{eq:cooft1}
2[L_{b,\gamma},\  t_1](\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))+L_{b,\gamma}(\dg_1,\dA_1)&=0,\\\label{eq:cooft0}
[[L_{b,\gamma},\  t_1],\ t_1](\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))+[L_{b,\gamma},\  t_1](\dg_1,\dA_1)+L_{b,\gamma}(\dg_0,\dA_0)&=0.
\end{align}

First, (<ref>) implies $L_{b,\gamma}\big(2t_1(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))+(\dg_1,\dA_1)\big)=0$, and thus by Proposition <ref>

\[
2t_1(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))+(\dg_1,\dA_1)=2(\hat{g}_{b_,s_1}(\sfS),\ \hat{A}_{b,s_1}(\sfS))+c(\dg_{b,s_1}(\sfS'),\ \dA_{b,s_1}(\sfS'))
\]

for some $c\in\BR$ and $\sfS'\in\BFS_1$. Subtracting $c(\dg_{b,s_1}(\sfS'),\ \dA_{b,s_1}(\sfS'))$ from $(\dg,\dA)$, we may assume $c=0$, and this implies

\[
(\dg_1,\dA_1)=2(\hat{g}_{b_,s_1}(\sfS),\ \hat{A}_{b,s_1}(\sfS))-2t_1(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS)).
\]

Plugging the above expression for $(\dg_1,\dA_1)$	into (<ref>) yields

\begin{equation}\label{eq:eqnforh0s1}
\begin{split}
L_{b,\gamma}(\dg_0,\dA_0)&=-[[L_{b,\gamma},\ t_1],\ t_1](\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))\\
&\quad-[L_{b,\gamma},\  t_1]\Big(2(\hat{g}_{b_,s_1}(\sfS),\ \hat{A}_{b,s_1}(\sfS))-2t_1(\dg_{b,s_1}(\sfS),\dA_{b,s_1}(\sfS))\Big) \in\eHb^{\infty,3/2-}(\CX)
\end{split}
\end{equation}

Our aim is to prove that there is no solution $(\dg_0,\dA_0)\in\eHb^{\infty,-1/2-}(\CX)$ to the above equation. To this end, we need to verify that the pairing of the right-hand side of (<ref>) with $(\dg^*_{b,s_1}(\sfS),\dA_{b,s_1}^*(\sfS))$ is non-zero. We note that it suffices to check the RN case $b=b_0$ because the general KN case with $b$ near $b_0$ follows directly by the continuity (in $b$). Using $\dA_{b_0,s_1}(\sfS)=\mathcal{O}_{\eHb^{\infty,1/2-}(\CX)}(\abs{\BQ_0})$ and the same argument as in the proof of Lemma <ref>, we obtain that the the pairing of the right-hand side of (<ref>) with $(\dg^*_{b_0,s_1}(\sfS),\dA_{b_0,s_1}^*(\sfS))$ is equal to

\begin{align*}
&\langle[[\Box_{g_{b_0},2},\  t_1],\ t_1]\dg_{b_0,s_1}(\sfS),\ \dg^*_{b_0,s_1}(\sfS)\rangle+2\langle[\Box_{g_{b_0},2},\  t_1](\hat{g}_{b_0,s_1}(\sfS)-t_1\dg_{b_0,s_1}(\sfS)),\  \dg^*_{b_0,s_1}(\sfS)\rangle+\mathcal{O}(\abs{\BQ_0}+\gamma)\\
\end{align*}

We first calculate $P_1$. Using $\dg_{b_0,s_1}(\sfS)=2\delta^*_{g_{b_0}}\omega_{b_0,s_1}(\sfS)$, $\dg^*_{b_0,s_1}(\sfS)=2G_{g_{b_0}}\delta^*_{g_{b_0}}\omega^*_{b_0,s_1}(\sfS)$ and (<ref>), we find that

\begin{align*}
P_1&=2\langle (\nabla^\al t_1\nabla_\al t_1)\dg_{b_0,s_1}(\sfS),\ \dg^*_{b_0,s_1}(\sfS)\rangle=2\langle 4(\weight-1)\dg_{b_0,s_1}(\sfS),\ \dg^*_{b_0,s_1}(\sfS)\rangle\\
&=2\langle16(\weight-1) \delta^*_{g_{b_0}}d\big((r-\Bm_0)\sfS\big),G_{g_{b_0}}\delta^*_{g_{b_0}}d\big((r-\Bm_0)H(r-\ehRN)\sfS\big)\rangle+\mathcal{O}(\abs{\BQ_0}^2+\gamma)\\
\end{align*}

Next, using $\hat{g}_{b_0,s_1}(\sfS)=2\delta_{g_{b_0}}^*\hat{\omega}_{b_0,s_1}(\sfS)$, (<ref>) and (<ref>), we compute

\begin{align*}
P_2&=2\langle\big(2\nabla^\al t_1\nabla_\al+\Box_{g_{b_0},0}t_1\big)(\hat{g}_{b_,s_1}(\sfS)-t_1\dg_{b,s_1}(\sfS)),\ \dg^*_{b_0,s_1}(\sfS)\rangle\\
&=-\frac{64}{9}(7-\log 8)\pi\Bm_0+\mathcal{O}(\abs{\BQ_0}^2+\gamma).
\end{align*}

Therefore, the pairing is indeed non-zero if $\abs{\BQ_0}+\gamma$ is sufficiently small.

§ STRUCTURE OF THE RESOLVENT OF THE LINEARIZED GAUGE-FIXED EINSTEIN-MAXWELL OPERATOR

In this section, we shall prove that $\widehat{L_{b,\gamma}}(\sigma)$ is invertible for $b=(\Bm,\Ba,\BQ)$ near $b_0=(\Bm_0,0,\BQ_0),\abs{\BQ_0}\ll\Bm_0$ and $\IM\sigma\geq 0, \sigma\neq 0$. This is a generalization of the first statement in Theorem <ref> from the RN case to the KN case. The proof uses the relevant conclusions of the RN case and the arguments in the proof of Proposition <ref>.

Let $\gamma>0$ be a fixed sufficiently small constant such that all the conclusions in the previous section hold. We recall the linearized gauge-fixed Einstein-Maxwell operator

\[
L_{b,\gamma}:=L_{(g_b,A_b),\gamma},\quad 	L^E_{b,\gamma}:=L^E_{(g_b,A_b),\gamma},\quad	L^M_{b,\gamma}:=L^M_{(g_b,A_b),\gamma},
\]

and its Fourier transformed version

\[
\widehat{L_{b,\gamma}}(\sigma):=e^{i\sigma t_{b,*}}L_{b,\gamma}e^{-i\sigma t_{b,*}}
\]
where $t_{b,*}=\chi_0(r)(t+r_{b_0,*})+(1-\chi_0(r))(t-r_{(\Bm,0,\BQ),*})$ is defined in (<ref>). Since $\gamma$ is fixed, we will drop the subscript $\gamma$ from the notations from now on.

Let $s>3$ and $-3/2<\ell<-1/2$. We redefine

\[
\mathcal{X}^{s,\ell}_{b}(\sigma):=\{(\dg,\dA)\in\eHb^{s,
\ell}(\CX;\scsym\oplus\scform): \widehat{L_{b}}(\sigma)(\dg,\dA)\in\eHb^{s-1,
\ell+2}(\CX;\scsym\oplus\scform)\}.
\]

As in the previous section, from now on, we will drop the notation of the bundle if it it clear from the context.

We also decompose zero energy nullspace $\mathcal{K}_b$ and the space of zero energy dual states $\mathcal{K}_b^*$ of $L_{b}$ as follows

\begin{equation}
\mathcal{K}_b=\mathcal{K}_{b,1}\oplus\mathcal{K}_{b,2},\quad \mathcal{K}^*_b=\mathcal{K}^*_{b,1}\oplus\mathcal{K}^*_{b,2}
\end{equation}

\begin{align}
\mathcal{K}_{b,1}:=\{(\dg_{b,s_1}(\BFS_1),\dA_{b,s_1}(\BFS_1))\},\quad \mathcal{K}_{b,2}:=\{(\dg_{b,s_0},\dA_{b,s_0})\}\oplus\{(\dg_{b,v_1}(\BFV_1),\dA_{b,v_1}(\BFV_1))\},\\
\mathcal{K}^*_{b,1}:=\{(\dg^*_{b,s_1}(\BFS_1),\dA^*_{b,s_1}(\BFS_1))\},\quad \mathcal{K}_{b,2}:=\{(\dg^*_{b,s_0},\dA^*_{b,s_0})\}\oplus\{(\dg^*_{b,v_1}(\BFV_1),\dA^*_{b,v_1}(\BFV_1))\}.
\end{align}

The motivation for doing such a decomposition is that $\widehat{L_{b}}(\sigma)$ is more singular when acting on $\mathcal{K}_{b,1}$ because of the existence of the generalized solutions to $L_{b}(\dg,\dA)=0$ with linear growth in $t_{b,*}$ and leading order terms in $\mathcal{K}_{b,1}$. Moreover, we recall the generalized zero energy states $(\hat{g}_{b,s_1}(\BFS_1),\hat{A}_{b,s_1}(\BFS_1))$ and dual states $(\hat{g}^*_{b,s_1}(\BFS_1),\hat{A}^*_{b,s_1}(\BFS_1))$ which is constructed in Proposition <ref> and define

\begin{equation}
\label{eq:stationaryofgezeromode}
\check{\mathcal{K}}_b:=\{(\check{g}_{b,s_1}(\sfS),\check{A}_{b,s_1}(\sfS)):\sfS\in\BFS_1\},\quad\check{\mathcal{K}}^*_b:=\{(\check{g}^*_{b,s_1}(\sfS),\check{A}^*_{b,s_1}(\sfS)):\sfS\in\BFS_1\}
\end{equation}

\begin{align}
\end{align}

§.§ Construction of the reference operator $\check{L}_{b}(\sigma)$.

We first construct a reference operator $\check{L}_{b}(\sigma)$ which is invertible near $\sigma=0$ by perturbing $\widehat{L_{b}}(\sigma)$ by a smoothing operator.

Let $b_0=(\Bm_0,0,\BQ_0)$ with $\abs{\BQ_0}\ll\Bm_0$.	There exist $V_b\in\Psi^{-\infty}(X;\scsym\oplus \scform)$ which is continuous in $b$ with uniformly compactly supported Schwartz kernel, and a constant $C_0>0$, such that the following statements hold.

* The operator
\[\check{L}_{b}(\sigma):=\widehat{L_{b}}(\sigma)+V_b:\mathcal{X}_b^{s,\ell}(\sigma)\to\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)
\]

is invertible for $\sigma\in\BC, \IM\sigma\geq 0$ with $\abs{\sigma}+\abs{b-b_0}\leq C_0$.

* The inverse operator $\check{L}_{b}(\sigma)^{-1}$ is continuous in $\sigma$ with values in $\mathcal{L}_{\mbox{weak}}(\eHb^{s-1,\ell+2}(\CX),\eHb^{s,\ell}(\CX))$ (the space of linear bounded operators equipped with weak operator topology), and  continuous in $\sigma$ with values in $\mathcal{L}_{\mbox{op}}(\eHb^{s-1+\epsilon,\ell+2+\epsilon}(\CX),\eHb^{s-\epsilon,\ell-\epsilon}(\CX))$ (norm topology) for any $\epsilon>0$.

One has

\begin{align}\label{eq:restriction1}
V_b(\check{g}_b,\check{A}_b)&=0\quad\mbox{for}\quad (\check{g}_b,\check{A}_b)\in\check{\mathcal{K}}_b.%\\\label{eq:restriction2}
%\langle(\dg,\dA),\left(I-(\check{L}_b(0)^{-1})^*V_b^*\right)(\check{g}^*_b,\check{A}^*_b)\rangle&=0\quad \mbox{for}\quad (\dg,\dA)\in\wt{\mathcal{K}}_b^\perp \quad  (\check{g}^*_b,\check{A}^*_b)\in\check{\mathcal{K}}^*_b.
\end{align}

The proof closely follows <cit.>. Concretely, we shall construct
\[V_b:\mathcal{D}'(X;\scsym\oplus\scform)\to C^\infty_c(X;S^2T^*X\oplus T^*X)
\]

where $V_{b}$ satisfy

\begin{align}\label{eq:requirement1}
\check{\mathcal{K}}_b\subset\ker V_{b},\quad
\Ran V_{b}\subset\mathcal{R}_b^\perp\quad\mbox{and}\quad	V_{b}|_{\mathcal{K}_b}:\mathcal{K}_b\to\mathcal{R}_b^\perp\quad\mbox{is invertible}.
%\check{\mathcal{K}}_b\in\ker V_{b,2},\quad\left(I-(\check{L}_b(0)^{-1})^*V_b^*\right)\check{\mathcal{K}}_b^*\cap \mbox{ann}(\wt{\mathcal{K}}_b)=0.
\end{align}

where $\mathcal{R}_b:=\left(\Ran_{\mathcal{X}_{b}^{s,\ell}(0)}\widehat{L_{b}}(0)\right)$. We now prove that once the above requirements are satisfied, all the three statements in the lemma hold. As for the proof of statements (1) and (2), it suffices to prove the invertiblity of $\check{L}_b(\sigma)=\widehat{L_{b}}(\sigma)+V_b$ for the case $b=b_0, \sigma=0$ since the invertibility and continuity for $(b,\sigma)$ near $(b_0,0)$ follow from the same arguments as in the proof of Lemma <ref>. We split the domain $\mathcal{X}_{b_0}^{s,\ell}(0)=\mathcal{K}_{b_0}^{\perp}\oplus\mathcal{K}_{b_0}$ and the target space $\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)=\mathcal{R}_{b_0}\oplus\mathcal{R}_{b_0}^\perp$. Under these splittings, $\check{L}_{b_0}(0)=\widehat{L_{b_0}}(0)+V_b$ takes the form

\[
\check{L}_{b_0}(0)=\begin{pmatrix}
\end{pmatrix}
\]

where $L_{00}=\widehat{L_{b_0}}(0)|_{\mathcal{K}_{b_0}^{\perp}}:\mathcal{K}_{b_0}^{\perp}\to\mathcal{R}_{b_0}$ is invertible. Then the invertibility of $\check{L}_{b_0}(0)$ follows

from that of  $V_{11}:\mathcal{K}_{b_0}\to\mathcal{R}_{b_0}^\perp$. We notice that the invertibility of $V_{11}$ holds provided that the above conditions in (<ref>) are satisfied.
For the statement (3), it is clear that the conclusion (<ref>) directly follows from the condition $\check{\mathcal{K}}_b\subset \ker V_{b}$.

First, since $(\dg_{b_0,s_1}(\BFS_1), \dA_{b_0,s_1}(\BFS_1))\in\eHb^{\infty,1/2-}(\CX)$ (see Theorem <ref>) and $(\check{g}_{b_0,s_1}(\BFS_1), \check{A}_{b_0,s_1}(\BFS_1))$ have nonzero $r^{-1}$ leading terms (see Lemma <ref>), we conclude that $\mathcal{K}_{b_0,1}\cap\check{\mathcal{K}}_{b_0}=0$, and thus $\mathcal{K}_{b_0}\cap\check{\mathcal{K}}_{b_0}=0$. By the continuity of $\mathcal{K}_b$ and $\check{\mathcal{K}}_b$ in $b$, this holds for $b$ near $b_0$ as well. This implies that fixing the basis $\{h_{b,1},\cdots, h_{b,8}\}$ of $\mathcal{K}_b$ and $\{\check{h}_{b,1},\check{h}_{b,2},\check{h}_{b,3}\}$ of $\check{\mathcal{K}}_b$, both of which depend continuously on $b$, by the same reasoning as in the beginning of the seventh step in the proof of Theorem <ref>, there exists $\{h^\flat_{b,1},\cdots, h^\flat_{b,8}\}\subset C^\infty_c(X^\circ;\scsym\oplus \scform)$ such that $\langle h_{b,i}, h^\flat_{b,j} \rangle=\delta_{ij}$ for $1\leq i,j \leq 8$ and $\langle\check{h}_{b,i}, h^\flat_{b,j}\rangle=0$ for $1\leq i\leq 3,1\leq j\leq 8$. (One can verify that $\{h^\flat_{b,1},\cdots, h^\flat_{b,8}\}\subset C^\infty_c(X^\circ)$ can be chosen to be continuous in $b$.) Next, we pick a basis of $\mathcal{R}_b^\perp$ in the following way. We note that $\mathcal{K}^*_b\cap\check{\mathcal{K}}^*_b=0$, which can be proved in a similar manner to the statement $\mathcal{K}_b\cap\check{\mathcal{K}}_b=0$. Fixing the continuous (in $b$) basis $\{h^*_{b,1},\cdots, h^*_{b,8}\}$ of $\mathcal{K}^*_b=\mbox{ann}(\mathcal{R}_b)$ and $\{\check{h}^*_{b,1}, \check{h}^*_{b,2}, \check{h}^*_{b,3}\}$ of $\check{\mathcal{K}}^*_b$, by the same reasoning mentioned above, there exists $\{h^\sharp_{b,1},\cdots, h^\sharp_{b,8}\}\subset C^\infty_c(X;\scsym\oplus \scform)$ (which can be chosen continuous in $b$) such that $\langle h^\sharp_{b,i}, h^*_{b,j}\rangle=\delta_{ij}$ for $1\leq i,j\leq 8$ and $\langle h^\sharp_{b,i}, \check{h}^*_{b,j}\rangle=0$ for $1\leq i\leq 8, 1\leq j\leq 3$. It is easy to check that $\{h^\sharp_{b,1},\cdots, h^\sharp_{b,8}\}$ are linearly independent and the space generated by $h^\sharp_{b,j}, 1\leq j\leq 8$ is a complement of $\mathcal{R}_b$. Therefore, we define

\[
V_{b}(h):=\sum_{i=1}^8h_{b,i}^\sharp\langle h, h_{b,i}^\flat\rangle,
\]

which satisfies the requirement (<ref>).

* Arrangement of (<ref>).

\[
V_{b,2}(h):=\sum_{i=1}^3v_{b,i}\langle h, h_{b,i}^\flat\rangle.
\]

It is clear that $\check{\mathcal{K}}\in\ker V_{b,2}$ and
\[V_b(\mathcal{K}_b)=\{h_{b,1}^\sharp+cv_{b,1},h_{b,2}^\sharp+cv_{b,2},h_{b,3}^\sharp+cv_{b,3},h_{b,4}^\sharp,h_{b,5}^\sharp,h_{b,6}^\sharp, h_{b,7}^\sharp\}\supset \{h_{b,4}^\sharp,\cdots, h_{b,7}^\sharp\}.
\]

As argued in the beginning, (<ref>) implies the invertibility of $\check{L}_{b}(0)$. Then after applying the invertible map $\check{L}_b(0)^*$, $\left(I-(\check{L}_b(0)^{-1})^*V_b^*\right)\check{\mathcal{K}}_b^*\cap \mbox{ ann}(\wt{\mathcal{K}}_b)=0$ is equivalent to

\begin{equation}\label{eq:requirement2equiv}
\widehat{L_{b}}(0)^*(\check{\mathcal{K}}_b^*)\cap
\end{equation}

Now we shall show that with the $V_{b,2}$ defined above, the following stronger condition holds.

\begin{equation}
\widehat{L_{b}}(0)^*(\check{\mathcal{K}}_b^*)\cap
\end{equation}

We notice that
\[\ann(\{h^\sharp_{b,4},\cdots,h_{b,7}^\sharp\})=\ker V_{b,1}^*\oplus \{h^*_{b,1}, \cdots, h^*_{b,3}\}=(u_b\oplus\check{\mathcal{K}}_b^*\ )\oplus \{h^*_{b,1}, \cdots, h^*_{b,3}\}
\]

where $U_b=\ann\{v_{b,1}, v_{b,2}, v_{b,3}\}\cap\ker V_{b,1}^*$ is a complement of $\check{\mathcal{K}}^*_b$ inside $\ker V_{b,1}^*$, and

\[
V_{b,2}^*(h^*)=\sum_{i=1}^3h^\flat_{b,i}\langle v_{b,i}, h^*\rangle.
\]

Then for any $h^*\in\ann(\{h^\sharp_{b,4},\cdots,h_{b,7}^\sharp\})$, we write it as $h^*=h_0^*+h_1^*+h_2^*$ with $h_0^*\in U_b, h_{1}^*\in \check{\mathcal{K}}_b^*$ and $h_{2}^*\in \{h^*_{b,1}, \cdots, h^*_{b,3}\}$, we have

\[
\widehat{L_{b}}(0)^*\omega^*=(\widehat{L_{b}}(0)+V_{b,1}+cV_{b,2})^*h^*=\widehat{L_{b}}(0)^*h_{0}^*+\Big(\widehat{L_{b}}(0)^*h_{1}^*+c\sum_{i=1}^3h^\flat_{b,i}\langle v_{b,i}, h_{1}^*\rangle\Big)+\sum_{i=1}^3h^\flat_{b,i}\langle h_{b,i}^\sharp, h^*_2\rangle
\]

we must have $h^*=\check{h}^*-c\sum_{i=1}^3h_{b,i}^*\langle v_{b,i}, \check{h}^*\rangle$

Using $\check{L}_{b}(\sigma)$ constructed in Lemma <ref>, we define the following spaces

\begin{equation}
\wt{\mathcal{K}}_{b,1}:=\check{L}_b(0)\mathcal{K}_{b,1},\quad	\wt{\mathcal{K}}_{b,2}:=\check{L}_b(0)\mathcal{K}_{b,2},\quad	\wt{\mathcal{K}}_{b}:=\check{L}_b(0)\mathcal{K}_b=	\wt{\mathcal{K}}_{b,1}\oplus	\wt{\mathcal{K}}_{b,2}.
\end{equation}

Since $\check{L}_{b}(0)\mathcal{K}_b=V_{b}(\mathcal{K}_b)$, we have $\wt{\mathcal{K}}_{b,1}, 	\wt{\mathcal{K}}_{b,2}, \wt{\mathcal{K}}_{b}\subset C^\infty_c(X;\scsym\oplus \scform)$.

Next, we decompose the target space $\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)$ into the range of $\widehat{L_{b}}(0)$ and its complement in a continuous (in $b$) way.

Let $b_0=(\Bm_0,0,\BQ_0)$ with $\abs{\BQ_0}\ll\Bm_0$. There exists a linear projection map
\[
\Pi_b^\perp:\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)\to\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform),
\]

which is of rank $8$ and depends continuously on $b$ near $b_0$ in the norm topology, such that

\[
\langle (I-\Pi_b^\perp)h,\  h^*\rangle=0\quad \mbox{for all}\quad h^*\in\mathcal{K}_b^*.
\]

Moreover, the $\Pi_{b}^\perp$ can be chosen such that it maps $\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)$ to $\C^\infty_c (X; \scsym\oplus \scform)$.

Let $\{h^*_{b,1}, \cdots,h^*_{b,8}\}$ be a basis which is continuous in $b$ of $\mathcal{K}^*_b$. There exists $\{h_{b,1}^\sharp, \cdots,h^\sharp_{b,8}\}\subset C^\infty_c(X;\scsym\oplus \scform)$ which is continuous in $b$, such that $\langle h_{b,i}^\sharp, h_{b,j}^*\rangle=\delta_{ij}$. We then define

\[
\Pi_b^\perp h=\sum_{i=1}^8h^\sharp_{b,i}\langle h, h^*_{b,i}\rangle.
\]

It is easy to check that $\Pi_b^\perp$ has rank $8$ and satisfies $	\langle (I-\Pi_b^\perp)h,\  h^*\rangle=0$ for all $h^*\in\mathcal{K}_b^*$.

We then define the projection onto $\mathcal{R}_b$ which is the range of $\widehat{L_{b}}(0)$

\[
\Pi_{b}=I-\Pi_{b}^\perp:\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)\to\mathcal{R}_b=\mbox{ann }\mathcal{K}_b^*.
\]

Now we are able to split the domain and target space of the operator $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}:\eHb^{s-1,\ell+2}\to\eHb^{s-1,\ell+2}$ as follows.

\begin{equation}\label{eq:splitofdoandran}
\begin{split}
\mbox{domain:}\quad \eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)=\wt{\mathcal{K}}_b^\perp\oplus\wt{\mathcal{K}}_{b,1}\oplus\wt{\mathcal{K}}_{b,2},\\
\mbox{target:}\quad \eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)=\mathcal{R}_b\oplus\mathcal{R}^\perp_{b,1}\oplus\mathcal{R}^\perp_{b,2}
\end{split}
\end{equation}

where $\mathcal{R}_{b,1}^\perp$, resp. $\mathcal{R}_{b,2}^\perp$ is a subspace of dimension $3$, resp. $5$ and defined as follows. Fixing the continuous (in $b$) basis $\{h^{*(1)}_{b,1},\cdots, h^{*(1)}_{b,3}\}$ of $\mathcal{K}^*_{b,1}$ and $\{h^{*(2)}_{b,1}, \cdots, h^{*(2)}_{b,5}\}$ of $\mathcal{K}^*_{b,2}$, there exist
\[\{v^{(1)}_{b,1},v^{(1)}_{b,2},v^{(1)}_{b,3}\},\   \{v^{(2)}_{b,1},v^{(2)}_{b,2}, v^{(2)}_{b,3},v^{(2)}_{b,4},v^{(2)}_{b,5}\}\subset C^\infty_c(X; \scsym\oplus \scform)
\]

such that $\langle v_{b,i}^{(1)}, h^{*(1)}_{b,j}\rangle=\delta_{ij}$ and $\langle v_{b,i}^{(2)}, h^{*(2)}_{b,j}\rangle=\delta_{ij}$. We then define

\[
\mathcal{R}_{b,1}^\perp=\{v^{(1)}_{b,1}, v^{(1)}_{b,2}, v^{(1)}_{b,3}\}\quad \mbox{and}\quad \mathcal{R}_{b,2}^\perp=\{v^{(2)}_{b,1}, v^{(2)}_{b,2}, v^{(2)}_{b,3} ,v^{(2)}_{b,4},v^{(2)}_{b,5}\}.
\]

As for the arrangement (11.39), it is equivalent to

\[
\langle \widehat{L_{b}}(0)\check{L}_b(0)^{-1}f, \check{h}^*\rangle=0 \quad\mbox{for}\quad f\in\widetilde{\mathcal{K}}_b^\perp,\  \check{h}^*\in\check{\mathcal{K}}^*_b.
\]

Since $\widehat{L_{b}}(0)\check{L}_b(0)^{-1}\widetilde{\mathcal{K}}_b^\perp$ is the range of $\widehat{L_{b}}(0)$, in order for the arrangement (11.39) to be true, $\check{h}^*$ is forced to lie in $\mathcal{K}^*_b$, which contradicts the fact $\mathcal{K}^*_b\cap\check{\mathcal{K}}_b^*=\{0\}$. Therefore, it seems that we are not able to arrange (11.39). Is this correct and if not could you tell me what I am missing?

However, as you said, the arrangement (11.39) could be omitted (while arrangement (11.38) is still needed to remove the singularity of the $R_{01}$ component of the inverse of $\widehat{L_{b}}(0)\check{L}_b(0)^{-1}$, is my understanding for arrangement (11.38) correct?). As you mentioned, without the arrangement (11.39), one needs to be more careful in low energy resolvent expansion. My understanding for this is as follows, and please let me know if there is anything I am missing.

Without arrangement (11.39), $\widehat{L_{b}}(0)\check{L}_b(0)^{-1}$ takes the form

\[
\widehat{L_{b}}(0)\check{L}_b(0)^{-1}=
\begin{pmatrix}
\widetilde{L}_{00}(\sigma)&\sigma^2\widetilde{L}_{01}(\sigma)&\sigma\widetilde{L}_{02}(\sigma)\\
\sigma\widetilde{L}_{20}(\sigma)&\sigma^2\widetilde{L}_{21}(\sigma)&\sigma\widetilde{L}_{22}(\sigma)
\end{pmatrix}
\]

where $\widetilde{L}_{00}, \widetilde{L}_{01},\widetilde{L}_{02}, \widetilde{L}_{20}$ are $\epsilon$-regular; $\widetilde{L}_{10},\widetilde{L}_{12},\widetilde{L}_{21},\widetilde{L}_{22}$ have an $\epsilon$-regular expansion up to order $1$, and $\widetilde{L}_{11}$ has an $\epsilon$-regular expansion up to order $2$. As a consequence, its inverse $R(\sigma)=\Big(\widehat{L_{b}}(0)\check{L}_b(0)^{-1}\Big)^{-1}$ is given by

\[
\begin{pmatrix}
\frac{\widetilde{R}_{10}(\sigma)}{\sigma}&\frac{\widetilde{R}_{11}(\sigma)}{\sigma^2}&\frac{\widetilde{R}_{12}(\sigma)}{\sigma}\\
\end{pmatrix}
\]

where $\widetilde{R}_{10}, \widetilde{R}_{12}$ are $\epsilon$-regular, and $\widetilde{R}_{11}, \widetilde{R}_{21},\widetilde{R}_{22}$ have an $\epsilon$-regular expansion up to order $1$. Therefore, we rewrite

\[
\begin{pmatrix}
\frac{\widetilde{R}_{10}(\sigma)}{\sigma}&\frac{\widetilde{R}^0_{11}+\sigma\widetilde{R}^e_{11}(\sigma)}{\sigma^2}&\frac{\widetilde{R}_{12}(\sigma)}{\sigma}\\
\end{pmatrix}.
\]

This implies that the solution $h$ to the equation $L_b h=f$ satisfies

\[
h=t_*h_s+\check{h}_s+h_v+h'_s+\tilde{h}+\tilde{h}'\quad\mbox{where}\quad h_s,h'_s\in\mathcal{K}_{b,s}, \check{h}_s\in\check{\mathcal{K}}_{b,s}, h_v\in\mathcal{K}_{b,v}.
\]

One of the remainder $\tilde{h}$ has the decay rate (in $t_*$) $t_*^{-1+\epsilon}$ and can be written as a sum of a pure gauge solution and a tail which decays  in $t_*$ at the rate $t_*^{-2+\epsilon}$. The other remainder $\tilde{h}'$ decays in $t_*$ at the rate $t_*^{-2+\epsilon}$.

§.§ Structure of the resolvent near zero energy

Now we are at the position of establishing the mode stability of the operator $\widehat{L_{b}}(\sigma)$ for $b$ near $b_0$ at $\IM\sigma\geq 0, \sigma\neq0$, and we also provide a description of the structure of the resolvent $\widehat{L_{b}}(\sigma)^{-1}$ near $\sigma=0$ along the way.

Let $(g_{b}, A_{b})$ be the RN metric and $4$-electromagnetic potential where $b=(\Bm, \Ba, \BQ)$ is near $b_0=(\Bm_0, 0,\BQ_0)$ with $\abs{\BQ_0}\ll\Bm_0$. Then the Fourier-transformed operator

\begin{equation}\label{eq:mlgEMmodenonzeroforRN}
\begin{split}
&\widehat{L_{b}}(\sigma):\{(\dg,\dA)\in\eHb^{s, \ell}(\CX;\scsym\oplus\scform): 	\widehat{L_{b}}(\sigma)(\dg,\dA)\in\eHb^{s,\ell+1}(\CX; \scsym\oplus\scform)\}\\
\end{split}
\end{equation}

is invertible for $\sigma\in\BC, \IM\sigma\geq 0, \sigma\neq 0$ and $s>3, \ell<-\frac12, s+\ell>-\frac 12$.

By the arguments in the last step of the proof of Theorem <ref> and Proposition <ref>, it suffices to prove that $\mathcal{X}^{s,\ell}_b(\sigma)\to\eHb^{s,\ell+1}(\CX;\scsym\oplus\scform)$ is invertible for $s>3, -3/2<\ell<-1/2$ and $(b,\sigma)$ with $\IM\sigma\geq 0, \sigma\neq 0$ near $(b_0,0)$. To this end, we shall prove the invertibility of the operator

\[
\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}:\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)\to\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)
\]

for $s>3, -3/2<\ell<-1/2$ and $(b,\sigma)$ with $\IM\sigma\geq 0, \sigma\neq 0$ near $(b_0,0)$.

Under the splitting (<ref>), we express

\begin{equation}\label{eq:exofLL-1}
\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}=
\begin{pmatrix}
\end{pmatrix}.
\end{equation}

It is clear that $L_{ij}(b,0)=0$ when $(i,j)\neq (0,0)$ and $L_{00}(b,0)$ is invertible. We will use the following resolvent identity (whose proof is similar to that of the resolvent identity (<ref>) in the proof of Lemma <ref>) frequently in the subsequent analysis

\begin{align}\label{eq:resolventidentity1}
\big(\check{L}_b(\sigma)^{-1}-\check{L}_b(0)^{-1}\big)(\dg,\dA)&=\check{L}_b(\sigma)^{-1}\big(\check{L}_b(0)-\check{L}_b(\sigma)\big)\check{L}_b(0)^{-1}(\dg,\dA)\\\label{eq:dualresolventidentity}			\big((\check{L}_b(\sigma)^{-1})^*-(\check{L}_b(0)^{-1})^*\big)(\dg^*,\dA^*)&=(\check{L}_b(\sigma)^{-1})^*\big(\check{L}_b(0)^*-\check{L}_b(\sigma)^*\big)(\check{L}_b(0)^{-1})^*(\dg^*,\dA^*)
\end{align}

for $\check{L}_b(0)^{-1}(\dg,\dA)\in \rho C^\infty(\CX)+\eHb^{s, 1/2-}(\CX)$ and $(\check{L}_b(0)^{-1})^*(\dg^*,\dA^*)\in \rho C^\infty(\CX)+\sHb^{-s+1, 1/2-}(\CX)$.

* Analysis of $L_{01}$ and $L_{02}$. For $(\tilde{g}_2, \tilde{A}_2)=\check{L}_b(0)(\dg_2,\dA_2)\in\wt{\mathcal{K}}_{b,2}$, we have

\begin{equation}\label{eq:calofL02}
\begin{split}
\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_2, \tilde{A}_2)&=\big(\check{L}_b(\sigma)-V_b\big)\check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg_2,\dA_2)\\
&=\sigma V_b\check{L}_b(\sigma)^{-1}\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma^2}{2}V_b\check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)
\end{split}
\end{equation}

where we use $\check{L}_b(0)^{-1}\check{L}_b(0)(\dg_2,\dA_2)=(\dg_2,\dA_2)\in \rho C^\infty(\CX)+\eHb^{\infty, 1/2-}(\CX)$ by Proposition <ref> and the resolvent identity (<ref>) in the third step. Since $(\dg_2,\dA_2)\in\rho C^\infty(\CX)+\eHb^{\infty,1/2-}(\CX)$ which is annihilated by normal operator of $\pa_\sigma\widehat{L_{b}}(0)$ modulo $\eHb^{\infty,3/2-}(\CX)$, it follows that
\[
\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)\in\eHb^{\infty,3/2-}(\CX).
\]

As $\pa^2_\sigma\widehat{L_{b}}(0)\in\rho^2C^\infty(\CX;\mbox{End}(\scsym\oplus\scform))$, we have
\[\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)\in\eHb^{\infty,3/2-}(\CX)
\]

as well. By using the continuity of $\check{L}_b(\sigma)^{-1}$ in the norm operator topology
\[\mathcal{L}_{\mbox{op}}(\eHb^{s-1+\epsilon, \ell+2+\epsilon}(\CX), \eHb^{s-\epsilon, \ell-\epsilon}(\CX)),
\]

we conclude that

\[
\check{L}_b(\sigma)^{-1}\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\ \check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)
\]

are continuous in $(b,\sigma)$, and thus
\[
\]

where $\wt{L}_{02}(b,\sigma):\wt{\mathcal{K}}_{b,2}\to\mathcal{R}_b$ is continuous in $(b,\sigma)$  in the norm operator topology.

For $(\tilde{g}_1, \tilde{A}_1)=\check{L}_b(0)(\dg_1,\dA_1)\in\wt{\mathcal{K}}_{b,1}$, the calculation in (<ref>) still applies. We further calculate

\begin{align*}
&\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_1, \tilde{A}_1)\\
&\quad=\sigma V_b\check{L}_b(\sigma)^{-1}\pa_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)+\frac{\sigma^2}{2}V_b\check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)\\
&\quad=i\sigma V_b\check{L}_b(\sigma)^{-1}\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{\sigma^2}{2}V_b\check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)\\
&\quad=i\sigma\big(I-V_b\check{L}_b(0)^{-1}\big)V_b(\check{g}_1,\check{A}_1)+i\sigma V_b(\check{L}_b(\sigma)^{-1}-\check{L}_b(0)^{-1})\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)\\
&\quad =i\sigma V_b(\check{L}_b(\sigma)^{-1}-\check{L}_b(0)^{-1})\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{\sigma^2}{2}V_b\check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)
\end{align*}

where we use Lemma <ref> in the last step. Since by Lemma <ref> and Lemma <ref>, we have
\[\check{L}_b(0)^{-1}\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)=(\check{g}_1,\check{A}_1)\in\rho C^\infty(\CX)+\eHb^{\infty,1/2-}(\CX),
\]

and it follows that we can apply the resolvent identity (<ref>) again and obtain

\begin{equation}\label{eq:calofL01}
\begin{split}
&\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_1, \tilde{A}_1)\\
&\quad=i\sigma V_b\check{L}_b(\sigma)^{-1}\big(\check{L}_b(0)-\check{L}_b(\sigma)\big)\check{L}_b(0)^{-1}\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{\sigma^2}{2}V_b\check{L}_b(\sigma)^{-1}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_1,\dA_1)\\
\end{split}
\end{equation}

By the same reasoning as in the component $L_{02}(b,\sigma)$, we find that

\[
\]

where $\wt{L}_{01}(b,\sigma):\wt{\mathcal{K}}_{b,1}\to\mathcal{R}_b$ is continuous in $(b,\sigma)$ in the norm operator topology.

* Analysis of $L_{10}$ and $L_{20}$. For $(\tilde{g}_0, \tilde{A}_0)\in\wt{\mathcal{K}}_{b}^\perp$ and $(\dg^*_{2}, \dA^*_2)\in\mathcal{K}^*_{b,2}$, we have

\begin{equation}\label{eq:calofL20}
\begin{split}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_0, \tilde{A}_0),\ (\dg^*_{2}, \dA^*_2)\rangle\\
&\quad=\langle(\tilde{g}_0, \tilde{A}_0),\ (\check{L}_b(\sigma)^{-1})^*\widehat{L_{b}}(\sigma)^*(\dg^*_{2}, \dA^*_2)\rangle\\
&\quad=\langle(\tilde{g}_0, \tilde{A}_0),(\dg^*_{2}, \dA^*_2)\rangle-\langle(\tilde{g}_0, \tilde{A}_0),(\check{L}_b(\sigma)^{-1})^*V_b^*(\dg^*_{2}, \dA^*_2)\rangle\\
&\quad=-\Big\langle(\tilde{g}_0, \tilde{A}_0),\big((\check{L}_b(\sigma)^{-1})^*-(\check{L}_b(0)^{-1})^*\big)V_b^*(\dg^*_{2}, \dA^*_2)\Big\rangle\\
&\quad=\Big\langle(\tilde{g}_0, \tilde{A}_0),(\check{L}_b(\sigma)^{-1})^*\big(\check{L}_b(\sigma)^*-\check{L}_b(0)^*\big)(\check{L}_b(0)^{-1})^*V_b^*(\dg^*_{2}, \dA^*_2)\Big\rangle\\
&\quad=\sigma\Big\langle (\tilde{g}_0, \tilde{A}_0), (\check{L}_b(\sigma)^{-1})^*\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)^*(\dg^*_2,\dA^*_2)\Big\rangle
\end{split}
\end{equation}

where we use $(\check{L}_b(0)^{-1})^*V_b^*(\dg^*_2,\dA^*_2)=(\dg^*_2,\dA^*_2)\in \rho C^\infty(\CX)+\sHb^{-3/2-C(a,\gamma), 1/2-}(\CX)$ where $C(a,\gamma)$ is a small constant by Proposition <ref> and the resolvent identity (<ref>) in the fourth step. Since $(\dg^*_2,\dA^*_2)$ is annihilated by normal operator of $(\pa_\sigma\widehat{L_{b}}(0))^*$ modulo $\sHb^{-5/2-C(a,\gamma),3/2-}(\CX)$, it follows that $(\pa_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)\in\sHb^{-5/2-C(a,\gamma),3/2-}(\CX)$. As $(\pa^2_\sigma\widehat{L_{b}}(0))^*\in\rho^2C^\infty(\CX)$, we have
\[(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)\in\sHb^{-3/2-\nu-,3/2-}(\CX).
\]

Since $(\check{L}_b(\sigma)^{-1})^*$ is continuous with values in $\mathcal{L}_{\mathrm{op}}(\sHb^{-s+\epsilon, -\ell+\epsilon}(\CX), \sHb^{-s+1-\epsilon, -\ell-2-\epsilon}(\CX))$, we conclude that

\[
(\check{L}_b(\sigma)^{-1})^*(\pa_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2),\ (\check{L}_b(\sigma)^{-1})^*(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)
\]

are continuous in $(b,\sigma)$, and thus
\[
\]

where $\wt{L}_{20}(b,\sigma):\wt{\mathcal{K}}^\perp_{b}\to\mathcal{R}^\perp_{b,2}$ is continuous in $(b,\sigma)$ in the norm operator topology.

For $(\tilde{g}_0, \tilde{A}_0)\in\wt{\mathcal{K}}_{b}^\perp$ and $(\dg^*_{1}, \dA^*_1)\in\mathcal{K}^*_{b,1}$, the calculation in (<ref>) still applies. We further calculate

\begin{align*}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_0, \tilde{A}_0),\ (\dg^*_1, \dA^*_1)\rangle\\
&\quad=\sigma\Big\langle (\tilde{g}_0, \tilde{A}_0), (\check{L}_b(\sigma)^{-1})^*\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0)\big)^*(\dg^*_1,\dA^*_1)\Big\rangle\\
&\quad=\sigma\Big\langle (\tilde{g}_0, \tilde{A}_0), i(\check{L}_b(\sigma)^{-1})^*\widehat{L_{b}}(0)^*(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad\quad+\sigma^2\Big\langle(\tilde{g}_0, \tilde{A}_0),\frac{1}{2}(\check{L}_b(\sigma)^{-1})^*(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_1,\dA^*_1)\Big\rangle\\
&\quad=\sigma\Big\langle (\tilde{g}_0, \tilde{A}_0), i\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad\quad+\sigma\Big\langle (\tilde{g}_0, \tilde{A}_0), i\big((\check{L}_b(\sigma)^{-1})^*-(\check{L}_b(0)^{-1})^*\big)\widehat{L_{b}}(0)^*(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad\quad\quad+\sigma^2\Big\langle(\tilde{g}_0, \tilde{A}_0),\frac{1}{2}(\check{L}_b(\sigma)^{-1})^*(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_1,\dA^*_1)\Big\rangle
\end{align*}

By a normal operator argument as in the proof of Proposition <ref>, we have

\[
(\check{L}_b(0)^{-1})^*: C^\infty_c(X)\to \rho C^\infty(\CX)+\sHb^{-3/2-C(a,\gamma),1/2-}(\CX).
\]

Therefore, by Lemma <ref>, we conclude
\[(\check{L}_b(0)^{-1})^*\widehat{L_{b}}(0)^*(\check{g}_1,\check{A}_1)=(\check{g}_1,\check{A}_1)-(\check{L}_b(0)^{-1})^*V^*_b(\check{g}_1,\check{A}_1)\in\rho C^\infty(\CX)+\sHb^{-3/2-C(a,\gamma),1/2-}(\CX).
\]

This implies that we can apply the resolvent identity (<ref>) again and obtain

\begin{equation}\label{eq:calofL10}
\begin{split}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_0, \tilde{A}_0),\ (\dg^*_1, \dA^*_1)\rangle\\
&\quad=\sigma\Big\langle (\tilde{g}_0, \tilde{A}_0), i\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_2,\check{A}^*_2)\Big\rangle\\
&\quad\quad-\sigma^2\Big\langle (\tilde{g}_0, \tilde{A}_0), i(\check{L}_b(\sigma)^{-1})^*(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_2,\check{A}^*_2)\Big\rangle\\
&\quad\quad\quad+\sigma^2\Big\langle(\tilde{g}_0, \tilde{A}_0),\frac{1}{2}(\check{L}_b(\sigma)^{-1})^*(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_2,\dA^*_2)\Big\rangle.
\end{split}
\end{equation}

By the same reasoning as in the component $L_{20}(b,\sigma)$, we find that

\[
L_{10}(b,\sigma)=\sigma \wt{L}_{10}(b,\sigma)
\]

where $\wt{L}_{10}(b,\sigma)=\wt{L}^0_{10}(b)+\sigma\wt{L}^e_{10}(b,\sigma)$ with $\wt{L}^e_{10}(b,\sigma):\wt{\mathcal{K}}_{b}^\perp\to \mathcal{R}_{b,1}^\perp$ continuous in $(b,\sigma)$ in the norm operator topology.

* Analysis of $L_{22}$. Using (<ref>), we find that

\begin{equation}\label{eq:calofL22}
\begin{split}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_2,\tilde{A}_2),\  (\dg_2^*, \dA_2^*)\rangle\\
&\quad=\langle\sigma\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma^2}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\  (\check{L}_b(\sigma)^{-1})^*V_b^*(\dg_2^*, \dA_2^*)\rangle\\
&\quad=\langle\sigma\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma^2}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\  (\dg_2^*, \dA_2^*)\rangle\\
&\quad\quad+\langle\sigma\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma^2}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\ \big((\check{L}_b(\sigma)^{-1})^*-(\check{L}_b(0)^{-1})^*\big)V_b^*(\dg_2^*, \dA_2^*)\rangle\\
&\quad=\langle\sigma\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma^2}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\  (\dg_2^*, \dA_2^*)\rangle\\
&\quad\quad+\langle\sigma\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma^2}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\ (\check{L}_b(\sigma)^{-1})^*\big(\check{L}_b(0)^*-\check{L}_b(\sigma)^*\big)(\dg_2^*, \dA_2^*)\rangle\\
&\quad=\sigma\langle-i[L_b, t_{b,*}](\dg_2, \dA_2)\,\  (\dg_2^*, \dA_2^*)\rangle+\sigma^2\Big(\langle\frac{1}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2),\  (\dg_2^*, \dA_2^*)\rangle\\
&\quad\quad-\big\langle\pa_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\dg_2,\dA_2), (\check{L}_b(\sigma)^{-1})^*(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg_2^*, \dA_2^*)\big\rangle\Big)
\end{split}
\end{equation}

where we use $(\check{L}_b(0)^{-1})^*V_b^*(\dg_2^*, \dA_2^*)=(\dg_2^*, \dA_2^*)$ and the resolvent identity (<ref>) in the third step. According to the proof of the non-existence of the linearly growing generalized zero modes with leading order term $(\dg_2, \dA_2)\in\mathcal{K}_{b,2}$ (see Lemma <ref> and <ref>), the pairing on $\mathcal{K}_{b,2}\times\mathcal{K}_{b,2}^*$

\[
\langle-i[L_b, t_{b,*}](\dg_2, \dA_2),\  (\dg_2^*, \dA_2^*)\rangle
\]

is non-degenerate for $b$ near $b_0$. Therefore, we find that

\[
L_{22}(b,\sigma)=\sigma\wt{L}_{22}(b,\sigma)\quad\mbox{with}\quad \wt{L}_{22}(b,\sigma)=\wt{L}^0_{22}(b)+\sigma\wt{L}^e_{22}(b,\sigma)
\]

where $\wt{L}^0_{22}(b)$ is invertible and $\wt{L}^e_{22}(b,\sigma):\wt{\mathcal{K}}_{b,2}\to\mathcal{R}_{b,2}^\perp$ is continuous in $(b,\sigma)$ in the norm operator topology.

* Analysis of $L_{11}$. Using the calculation in (<ref>) and (<ref>), we see that

\begin{equation}\label{eq:calofL11}
\begin{split}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_1,\tilde{A}_1),\  (\dg_1^*, \dA_1^*)\rangle\\
&\ =\sigma^2\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\check{L}_b(\sigma)^{-1})^*V_b^*(\dg_1^*, \dA_1^*)\rangle\\
&\ =\sigma^2\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\dg_1^*, \dA_1^*)\rangle\\
&\ \ +\sigma^2\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\check{L}_b(\sigma)^{-1})^*(\check{L}_{b}(0)-\check{L}_{b}(\sigma))^*(\dg_1^*, \dA_1^*)\rangle\\
&\ =\sigma^2\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\dg_1^*, \dA_1^*)\rangle\\
&\ \ -\sigma^3\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  i(\check{L}_b(\sigma)^{-1})^*\widehat{L_{b}}(0)^*(\check{g}_1^*, \check{A}_1^*)\rangle\\
&\ \ \  -\sigma^4\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\check{L}_b(\sigma)^{-1})^*(\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0))^*(\dg_1^*, \dA_1^*)\rangle	\\
&\ =\sigma^2\langle-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\dg_2^*, \dA_2^*)\rangle+\sigma^3\langle\frac{-i}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1), (\dg_1^*,\dA_1^*)\rangle\\
&\ \ -\sigma^3\langle-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  i(I-(\check{L}_b(0)^{-1})^*V_b^*)(\check{g}_1^*, \check{A}_1^*)\rangle\\
&\ \ \ -\sigma^4\langle\frac{-i}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1),\  i(I-(\check{L}_b(0)^{-1})^*V_b^*)(\check{g}_1^*, \check{A}_1^*)\rangle\\
&\ \ \ \  +\sigma^4\Big\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\\
&\quad\quad\quad\quad\quad\quad (\check{L}_b(\sigma)^{-1})^*(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0))^*(I-(\check{L}_b(0)^{-1})^*V_b^*)(\check{g}_1^*, \check{A}_1^*)\Big\rangle	\\
&\ \ \ \ \ \  -\sigma^4\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\check{L}_b(\sigma)^{-1})^*(\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0))^*(\dg_1^*, \dA_1^*)\rangle	.
\end{split}
\end{equation}

According to the proof of the non-existence of the quadratically growing generalized zero modes with leading order term $(\dg_1, \dA_1)\in\mathcal{K}_{b,1}$ (see Lemma <ref>), the pairing on $\mathcal{K}_{b,1}\times\mathcal{K}_{b,1}^*$

\begin{align*}
&\langle-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\  (\dg_2^*, \dA_2^*)\rangle\\
&\quad=-\langle[L_b, t_{b,*}](\check{g}_1,\check{A}_1)+\frac{1}{2}[[L_b, t_{b,*}],t_{b,*}](\dg_1,\dA_1),\  (\dg_2^*, \dA_2^*)\rangle
\end{align*}

is non-degenerate for $b$ near $b_0$. Therefore, we find that

\[
L_{11}(b,\sigma)=\sigma^2\wt{L}_{11}(b,\sigma)\quad\mbox{with}\quad \wt{L}_{11}(b,\sigma)=\wt{L}^0_{11}(b)+\sigma\wt{L}^1_{22}(b)+\sigma^2\wt{L}^e_{11}(b,\sigma)
\]

where $\wt{L}^0_{11}(b)$ is invertible and $\wt{L}^e_{11}(b,\sigma):\wt{\mathcal{K}}_{b,1}\to\mathcal{R}_{b,1}^\perp$ is continuous in $(b,\sigma)$ in the norm operator topology.

* Analysis of $L_{12}$ and $L_{21}$. For $(\tilde{g}_1,\tilde{A}_1)\in\wt{\mathcal{K}}_{b,1}$ and $(\dg_2^*, \dA_2^*)\in\mathcal{K}_{b,2}^*$, using (<ref>) and (<ref>), we compute

\begin{equation}\label{eq:calofL21}
\begin{split}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_1,\tilde{A}_1),\  (\dg_2^*, \dA_2^*)\rangle\\
&\ =\sigma^2\langle-i\pa_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\ (\dg_2^*, \dA_2^*)\rangle+\sigma^3\langle\frac{-i}{2}\pa^2_\sigma\widehat{L_{b}}(0)(\check{g}_1,\check{A}_1), (\dg_2^*,\dA_2^*)\rangle\\
&\ \ -\sigma^3\Big\langle-i\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)(\check{g}_1,\check{A}_1)+\frac{1}{2}\pa_\sigma^2\widehat{L_{b}}(0)(\dg_1,\dA_1),\\
&\qquad\qquad\quad\  (\check{L}_b(\sigma)^{-1})^*\big(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa^2_\sigma\widehat{L_{b}}(0)\big)^*(\dg_2^*, \dA_2^*)\Big\rangle.
\end{split}
\end{equation}

\[
L_{21}(b,\sigma)=\sigma^2\wt{L}_{21}(b,\sigma)\quad\mbox{with}\quad \wt{L}_{21}(b,\sigma)=\wt{L}^0_{21}(b)+\sigma\wt{L}^e_{21}(b,\sigma)
\]

where $\wt{L}^e_{21}(b,\sigma):\wt{\mathcal{K}}_{b,1}\to\mathcal{R}^\perp_{b,2}$ is continuous in $(b,\sigma)$ in the norm operator topology.

For $(\tilde{g}_2,\tilde{A}_2)=\check{L}_b(0)(\dg_2, \dA_2)\in\wt{\mathcal{K}}_{b,2}=\check{L}_b(0)\mathcal{K}_{b,2}$ and $(\dg_1^*, \dA_1^*)\in\mathcal{K}^*_{b,1}$, using (<ref>), we calculate

\begin{equation}\label{eq:calofL12}
\begin{split}
&\langle\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_2, \tilde{A}_2),\ (\dg^*_1, \dA^*_1)\rangle\\
&\quad=\sigma\Big\langle \big(I-V_b(\check{L}_b(0)^{-1})\big)\check{L}_b(0)(\dg_2, \dA_2), i(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad\quad-\sigma^2\Big\langle \check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg_2, \dA_2), i(\pa_\sigma\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad\quad\quad+\sigma^2\Big\langle\check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg_2, \dA_2),\frac{1}{2}(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_1,\dA^*_1)\Big\rangle\\
&\quad\quad\quad\quad-\sigma^3\Big\langle \check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg_2, \dA_2), \frac{i}{2}(\pa_\sigma\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad=-\sigma^2\Big\langle (\dg_2, \dA_2), i(\pa_\sigma\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_2,\check{A}^*_2)-\frac{1}{2}(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_1,\dA^*_1)\Big\rangle\\
&\quad\quad+\sigma^3\Big\langle \check{L}_b(\sigma)^{-1}(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0))(\dg_2, \dA_2), i(\pa_\sigma\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_1,\check{A}^*_1)\Big\rangle\\
&\quad\quad\quad+\sigma^3\Big\langle \check{L}_b(\sigma)^{-1}(\pa_\sigma\widehat{L_{b}}(0)+\frac{\sigma}{2}\pa_\sigma^2\widehat{L_{b}}(0))(\dg_2, \dA_2), -\frac{1}{2}(\pa^2_\sigma\widehat{L_{b}}(0))^*(\dg^*_1,\dA^*_1)\Big\rangle\\
&\quad\quad\quad\quad-\sigma^3\Big\langle \check{L}_b(\sigma)^{-1}\check{L}_b(0)(\dg_2, \dA_2), \frac{i}{2}(\pa^2_\sigma\widehat{L_{b}}(0))^*\big(I-(\check{L}_b(0)^{-1})^*V_b^*\big)(\check{g}^*_1,\check{A}^*_1)\Big\rangle.
\end{split}
\end{equation}

\[
L_{12}(b,\sigma)=\sigma^2\wt{L}_{12}(b,\sigma)\quad\mbox{with}\quad \wt{L}_{12}(b,\sigma)=\wt{L}^0_{12}(b)+\sigma\wt{L}^e_{12}(b,\sigma)
\]

where $\wt{L}^e_{12}(b,\sigma):\wt{\mathcal{K}}_{b,2}\to\mathcal{R}^\perp_{b,1}$ is continuous in $(b,\sigma)$ in the norm operator topology.

* Analysis of $L_{00}$.
Following the arguments in the first step in the proof of Lemma <ref>, we can prove that for $(b,\sigma)$ near $(b,0)$, $L_{00}(b,\sigma)$ is invertible and there exists a uniform constant $C>0$ such that

\begin{equation}
\norm{L_{00}(b,\sigma)^{-1}}_{\mathcal{R}_b\to\wt{\mathcal{K}}_{b}^\perp}\leq C.
\end{equation}

Moreover, for $(\tilde{g}_0, \tilde{A}_0)\in\wt{\mathcal{K}}_{b}^\perp$, we write

\begin{equation}\label{eq:calofL00}
\begin{split}
L_{00}(b,\sigma)(\tilde{g}_0, \tilde{A}_0)&=\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}(\tilde{g}_0, \tilde{A}_0)-\big(L_{10}(b,\sigma)+L_{20}(b,\sigma)\big)(\tilde{g}_0, \tilde{A}_0)\\
&=(\tilde{g}_0, \tilde{A}_0)-V_b\check{L}_b(\sigma)^{-1}(\tilde{g}_0, \tilde{A}_0)-\big(L_{10}(b,\sigma)+L_{20}(b,\sigma)\big)(\tilde{g}_0, \tilde{A}_0).
\end{split}
\end{equation}

Since $\check{L}_b(\sigma)^{-1}$ is continuous in the norm operator topology $\mathcal{L}_{\mathrm{op}}(\eHb^{s-1+\epsilon, \ell+2+\epsilon}(\CX), \eHb^{s-\epsilon, \ell-\epsilon}(\CX))$ and $V_b$ maps $\mathcal{D}'(X^\circ)$ to $C^\infty_c(X)$, it follows that $L_{00}(b,\sigma):\wt{\mathcal{K}}_b^\perp\to\mathcal{R}_b$ is continuous in $(b,\sigma)$ in the norm operator topology.

* The inverse of $\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}$. Putting all the above analysis of the components $L_{ij}(b,\sigma)$ together, we conclude that

\begin{equation}\label{eq:detailedexofLL-1}
\begin{split}
L_{00}(b,\sigma)& \sigma^2\wt{L}_{01}(b,\sigma)&\sigma\wt{L}_{02}(b,\sigma)\\
\sigma\wt{L}_{10}(b,\sigma)& \sigma^2\wt{L}_{11}(b,\sigma)&\sigma^2\wt{L}_{12}(b,\sigma)\\
\sigma\wt{L}_{20}(b,\sigma)& \sigma^2\wt{L}_{21}(b,\sigma)&\sigma\wt{L}_{22}(b,\sigma)\\
\end{pmatrix}\\
L_{00}(b,\sigma)& \sigma^2\wt{L}_{01}(b,\sigma)&\sigma\wt{L}_{02}(b,\sigma)\\
\sigma\big(\wt{L}^0_{10}(b)+\sigma\wt{L}^e_{10}(b,\sigma)\big)&\sigma^2\big(\wt{L}^0_{11}(b)+\sigma\wt{L}^1_{11}(b)+\sigma^2\wt{L}^e_{11}(b,\sigma)\big)&\sigma^2\big(\wt{L}^0_{12}(b)+\sigma\wt{L}^e_{12}(b,\sigma)\big)\\
\sigma\wt{L}_{20}(b,\sigma)&\sigma^2\big(\wt{L}^0_{21}(b)+\sigma\wt{L}^e_{21}(b)\big)&\sigma\big(\wt{L}^0_{22}(b)+\sigma\wt{L}^e_{22}(b,\sigma)\big)
\end{pmatrix}
\end{split}
\end{equation}

where $L_{00}(b,\sigma), \wt{L}^0_{11}(b), \wt{L}^0_{22}(b)$ are invertible, and $L_{00}(b,\sigma)$, $\wt{L}_{ij}(b,\sigma)$ with $(i,j)=(0,1), (0,2), (2,0)$, $\wt{L}^e_{ij}(b,\sigma)$ with $(i,j)=(1,0), (1,1), (1,2), (2,1), (2,2)$ are continuous in $(b,\sigma)$ in the norm operator topology.

Solving the system

\[
\widehat{L_{b}}(\sigma)\check{L}_b(\sigma)^{-1}\big((\tilde{g}_0,\tilde{A}_0), (\tilde{g}_1,\tilde{A}_1),(\tilde{g}_2,\tilde{A}_2)\big)=(f_0, f_1,f_2)
\]

\begin{equation}\label{eq:exofinverseR}
\begin{pmatrix}
\wt{R}_{00}(b,\sigma)&\wt{R}_{01}(b,\sigma)&\wt{R}_{02}(b,\sigma)\\
\sigma^{-1}\wt{R}_{10}(b,\sigma)&\sigma^{-2}\wt{R}_{11}(b,\sigma)&\sigma^{-1}\wt{R}_{12}(b,\sigma)\\
\wt{R}_{20}(b,\sigma)&\sigma^{-1}\wt{R}_{21}(b,\sigma)&\sigma^{-1}\wt{R}_{22}(b,\sigma)
\end{pmatrix}
\end{equation}

\begin{equation}\label{eq:exofcompoofR}
\begin{split}
\wt{R}_{11}=\big(\wt{L}^\sharp_{11}-\sigma\wt{L}^\sharp_{12}(\wt{L}^\flat_{22})^{-1}\wt{L}^\sharp_{21}\big)^{-1},\quad \wt{R}_{10}=\wt{R}_{11}\big(-\wt{L}_{10}+\sigma\wt{L}_{12}^\sharp(\wt{L}^\flat_{22})^{-1}\wt{L}_{20}\big)L_{00}^{-1},\\
\wt{R}_{12}=-\wt{R}_{11}\wt{L}_{12}^\sharp(\wt{L}^\flat_{22})^{-1},\quad \wt{R}_{21}=-(\wt{L}^\flat_{22})^{-1}\wt{L}_{21}^\sharp\wt{R}_{11},\quad\wt{R}_{22}=(\wt{L}^\flat_{22})^{-1}-\sigma(\wt{L}^\flat_{22})^{-1}\wt{L}_{21}^\sharp \wt{R}_{12},\\
\wt{R}_{20}=-(\wt{L}^\flat_{22})^{-1}\big(\wt{L}_{20}L_{00}^{-1}+\wt{L}_{21}^\sharp\wt{R}_{10}\big),\quad \wt{R}_{01}=-L_{00}^{-1}\big(\wt{L}_{01}\wt{R}_{11}+\wt{L}_{02}\wt{R}_{21}\big),\\
\wt{R}_{02}=-L_{00}^{-1}\big(\sigma\wt{L}_{01}\wt{R}_{12}+\wt{L}_{02}\wt{R}_{22}\big),\quad \wt{R}_{00}=L_{00}^{-1}\big(I-\sigma\wt{L}_{01}\wt{R}_{10}-\sigma\wt{L}_{02}\wt{R}_{20}\big)
\end{split}
\end{equation}

\begin{equation}\label{eq:exofauxofR}
\begin{split}
&\wt{L}^\sharp_{i1}=\wt{L}_{i1}-\sigma\wt{L}_{i0}L_{00}^{-1}\wt{L}_{01}\quad\mbox{for}\quad i=1,2, \quad \wt{L}^\sharp_{12}=\wt{L}_{12}-\wt{L}_{10}L_{00}^{-1}\wt{L}_{02},\\
&\qquad\qquad\qquad\qquad	\wt{L}_{22}^\flat=\wt{L}_{22}-\sigma\wt{L}_{20}L_{00}^{-1}\wt{L}_{02}.
\end{split}
\end{equation}

Since $\check{L}_b(\sigma)$ is invertible for $(b,\sigma)$ near $(b_0,0)$, it follows that $\widehat{L_{b}}(\sigma)^{-1}=\check{L}_b(\sigma)^{-1} R(b,\sigma)$, which implies the invertibility of $\widehat{L_{b}}(\sigma):\mathcal{X}_b^{s,\ell}\to\eHb^{s-1,\ell+1}(\CX)$ for $(b,\sigma)$ with $\IM\sigma\geq 0, \sigma\neq 0$ near $(b_0, 0)$.
This finishes the proof of the theorem.

We now give a more detailed description of the structure of the singular part of $\widehat{L_{b}}(\sigma)^{-1}$.

For $(b,\sigma)$ with $\IM\sigma\geq 0$ near $(b_0,0)$, we have

\[
\widehat{L_{b}}(\sigma)^{-1}=P(b,\sigma)+L_b^-(\sigma):\eHb^{s-1,\ell+2}(\CX;\scsym\oplus\scform)\to\eHb^{s,\ell}(\CX;\scsym\oplus\scform).
\]

The regular part $L^-_b(\sigma)$ is uniformly bounded, and is continuous in $\sigma$ with values in
\[\mathcal{L}_{\mathrm{weak}}(\eHb^{s-1,\ell+2}(\CX),\eHb^{s,\ell}(\CX))\cap\mathcal{L}_{\mathrm{op}}(\eHb^{s-1+\epsilon,\ell+2+\epsilon}(\CX),\eHb^{s-\epsilon,\ell-\epsilon}(\CX))
\]

for $\epsilon>0$.

The singular part $P(b,\sigma)$ satisfies

\begin{equation}
P(b,\sigma)f=(\sigma^{-2}(\dg_1,\dA_1)-i\sigma(\check{g}_1, \check{A}_1))+i\sigma^{-1}\big((\dg_2, \dA_2)
+(\dg'_1, \dA'_1)\big)+i\sigma^{-1}(\dg_1''(\sigma), \dA_1''(\sigma))
\end{equation}

\[
(\dg_1,\dA_1),\  (\dg_1',\dA_1'),\ (\dg_1''(\sigma),\dA_1''(\sigma))\in\mathcal{K}_{b,1},\quad (\check{g}_1, \check{A}_1)\in \check{\mathcal{K}}_b
\]

with $(\dg_1''(\sigma),\dA_1''(\sigma))=o(1)$ in $\mathcal{K}_{b,1}$ as $\sigma\to 0$, and
\[
\]

Moreover, $(\dg_1,\dA_1), (\dg_1',\dA_1')$ and $(\dg_2,\dA_2)$ are determined by the conditions

\begin{align}\notag