First, we prove the surjectivity of the linear operator $\mathscr{L}_{g}$. By standard Hodge theory, it holds that

(9.105)		$\displaystyle\Omega^{2}_{+}(\mathcal{M})$	$\displaystyle=\mathcal{H}^{+}(\mathcal{M})\oplus d^{+}(\Omega^{1}(\mathcal{M}))$
(9.106)		$\displaystyle\Omega^{1}(\mathcal{M})$	$\displaystyle=d(\Omega^{0}(\mathcal{M}))\oplus\mathring{\Omega}^{1}(\mathcal{M}),$

where $\mathring{\Omega}^{1}(\mathcal{M})$ denotes the space of divergence-free $1$-forms on $\mathcal{M}$, therefore

(9.107)

$$\Omega^{2}_{+}(\mathcal{M})=\mathcal{H}^{+}(\mathcal{M})\oplus d^{+}(\mathring{\Omega}^{1}(\mathcal{M})).$$

This clearly implies that

(9.108)

$$\mathscr{L}_{g}=(d^{+}\oplus\Id)\otimes\mathbb{R}^{3}:\mathfrak{A}\longrightarrow\mathfrak{B}.$$

is surjective.

The remainder of the proof is a contradiction argument. We will argue on the level of forms, and this will imply the result for triples. If (9.104) does not hold for a uniform constant, then there exists a sequence of gluing parameters $\beta_{j}\rightarrow\infty$ and $\eta_{j}$, $\bar{\xi}^{+}_{j}$ with

(9.109)		$\displaystyle e^{10\delta\cdot\beta_{j}}\|d^{+}\eta_{j}+\bar{\xi}^{+}_{j}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}$	$\displaystyle\rightarrow 0,$
(9.110)		$\displaystyle\|\eta_{j}\|_{C_{\delta,\nu,\mu}^{1,\alpha}(\mathcal{M})}+\|\bar{{\xi}}^{+}_{j}\|_{L^{2}(\mathcal{M})}$	$\displaystyle=1,$

as $j\to\infty$. Pairing $d^{+}\eta_{j}+\bar{\xi}^{+}_{j}$ with $\bar{\xi}^{+}_{j}$ and integrating, and using (9.109), we obtain that

(9.111)

$\displaystyle\begin{split}\|\bar{\xi}^{+}_{j}\|_{L^{2}(\mathcal{M})}^{2}&\leq\epsilon_{j}e^{-10\delta\cdot\beta_{j}}\int_{\mathcal{M}}|\bar{\xi}_{j}^{+}|(\rho_{\delta,\nu+1,\mu}^{(0+\alpha)})^{-1}\dvol_{g_{\beta_{j}}}\\ &\leq\epsilon_{j}e^{-10\delta\cdot\beta_{j}}\|\bar{\xi}^{+}_{j}\|_{L^{2}(\mathcal{M})}\Big\{\int_{\mathcal{M}}(\rho_{\delta,\nu+1,\mu}^{(0+\alpha)})^{-2}\dvol_{g_{\beta_{j}}}\Big\}^{\frac{1}{2}},\end{split}$

where $\epsilon_{j}\to 0$ as $j\to\infty$. It is easy to check that

(9.112)

$\displaystyle\int_{\mathcal{M}}(\rho_{\delta,\nu+1,\mu}^{(0+\alpha)})^{-2}\dvol_{g_{\beta_{j}}}<C,$

where $C$ is independent of $\beta$, so this implies that

(9.113)

$$e^{10\delta\cdot\beta_{j}}\|\bar{\xi}^{+}_{j}\|_{L^{2}(\mathcal{M})}\to 0$$

as $j\to\infty$.

Next, since the triple $\bm{\omega}^{\mathcal{M}}$ is harmonic and spans $\mathcal{H}_{+}(\mathcal{M})$ at every point, we can write

(9.114)

$$\bar{\xi}^{+}=\lambda_{1}\omega_{1}+\lambda_{2}\omega_{2}+\lambda_{3}\omega_{3}.$$

Recall by the definition of the triple $\bm{\omega}_{\beta}^{\mathcal{M}}$, for every $1\leq p,q\leq 3$,

(9.115)

$$\frac{1}{2}\int_{\mathcal{M}}\omega_{p}\wedge\omega_{q}=\int_{\mathcal{M}}Q_{pq}\dvol_{\bm{\omega}_{\beta}^{\mathcal{M}}},$$

and so for any self-dual harmonic form $\bar{\xi}^{+}\in\mathcal{H}_{+}(\mathcal{M})$,

(9.116)

$\displaystyle\|\bar{\xi}^{+}\|_{L^{2}(\mathcal{M})}^{2}$

$\displaystyle=2\sum_{p,q=1}^{3}\lambda_{p}\lambda_{q}\int_{\mathcal{M}}Q_{pq}\dvol_{\bm{\omega}_{\beta}^{\mathcal{M}}},$

so applying the volume estimate

(9.117)

$$C^{-1}\beta^{2}\leq\Vol_{g}(\mathcal{M})\leq C\beta^{2},$$

and Proposition 6.4, we have the estimate

(9.118)

$\displaystyle C^{-1}\beta_{j}^{2}(\lambda_{1,j}^{2}+\lambda_{2,j}^{2}+\lambda_{3,j}^{2})\leq\|\bar{\xi}_{j}^{+}\|_{L^{2}(\mathcal{M})}^{2}.$

The above and (9.113) imply that $\beta_{j}\lambda_{k,j}e^{10\delta\cdot\beta_{j}}\to 0$ as $j\to\infty$ for $k=1,2,3$. We then have

(9.119)

$\displaystyle\begin{split}\|\bar{\xi}^{+}_{j}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}&=\|\lambda_{1,j}\omega_{1}+\lambda_{2,j}\omega_{2}+\lambda_{3,j}\omega_{3}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}\\ &\leq\lambda_{1,j}\|\omega_{1}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}+\lambda_{2,j}\|\omega_{2}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}+\lambda_{3,j}\|\omega_{3}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}.\end{split}$

Since

(9.120)

$\displaystyle\|\omega_{k}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}\leq Ce^{5\delta\cdot\beta_{j}},$

for $1\leq k\leq 3$, the above implies that

(9.121)

$\displaystyle\|\bar{\xi}^{+}_{j}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}\leq C\epsilon_{j}\beta_{j}^{-1}e^{-5\delta\cdot\beta_{j}},$

for some sequence $\epsilon_{j}\to 0$ as $j\to\infty$, so we have proved that

(9.122)

$\displaystyle\|\bar{\xi}^{+}_{j}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}\rightarrow 0,$

as $j\to\infty$. Consequently, our sequence satisfies

(9.123)		$\displaystyle\|d^{+}\eta_{j}\|_{C_{\delta,\nu+1,\mu}^{0,\alpha}(\mathcal{M})}$	$\displaystyle\rightarrow 0,$
(9.124)		$\displaystyle\|\eta_{j}\|_{C_{\delta,\nu,\mu}^{1,\alpha}(\mathcal{M})}$	$\displaystyle\to 1,$

as $j\to\infty$, which contradicts Proposition 9.2. ∎