We just deal with the negative end of the neck, the positive end is similar. Let

(6.25)

$\displaystyle U_{-}=\{p\in(\mathbb{T}^{2}\times\mathbb{R})\setminus\{p_{1},\dots,p_{m_{0}}\}\ |\ -\infty<z<-T_{-}/2\}.$

Note that $U_{-}$ deformation retracts to $\mathbb{T}^{2}$, so $H^{2}(U_{-},\mathbb{Z})=\mathbb{Z}$. The neck is a circle bundle over $U=(\mathbb{T}^{2}\times\mathbb{R})\setminus\{p_{1},\dots,p_{m_{0}}\}$, and call the restriction to $U_{-}$ by $\mathcal{N}_{U_{-}}$. Note this bundle has Euler number $b_{-}$.

Over $U_{-}$ there exists another $S^{1}$-bundle explicitly identified with an open subset of the model space

(6.26)

$\displaystyle\mathcal{N}_{b_{-}}=(-\infty,-T_{-}+1,)\times\Nil^{3}_{b_{-}}(\epsilon,\tau),$

with connection form $\theta_{b_{-}}$, which has curvature form $-\frac{2\pi b_{-}}{A}dx\wedge dy$, so this bundle also has Euler number $b_{-}$. From the exponential sheaf sequence, $H^{1}(U_{-},\mathcal{E}^{*})\cong H^{2}(U_{-},\mathbb{Z})$, so there exists a bundle equivalence

(6.27)

$\displaystyle H:\mathcal{N}_{b_{-}}\rightarrow\mathcal{N}_{U_{-}}$

which covers the identity map on the base, and such that the pullback bundle $H^{*}\mathcal{N}_{U_{-}}=\mathcal{N}_{b_{-}}$. The $1$-forms $H^{*}\theta$ and $\theta_{b_{-}}$ are therefore both connection forms on $\mathcal{N}_{b_{-}}$. Note that

(6.28)

$\displaystyle d(H^{*}\theta)=H^{*}d\theta=H^{*}(*dV)=*dV_{\beta},$

and

(6.29)

$\displaystyle d\theta_{b_{-}}=-\frac{2\pi b_{-}}{A}dx\wedge dy.$

From the asymptotics on $V_{\beta}$ in (6.14), we have

(6.30)

$\displaystyle d(H^{*}\theta-\theta_{b_{-}})=O(e^{\delta^{\prime}z}),$

as $z\rightarrow-\infty$. By same method from the proof of Lemma 3.7, we conclude that

(6.31)

$\displaystyle d(H^{*}\theta-\theta_{b_{-}})=da,$

where $a=O(e^{\delta^{\prime}z})$, as $z\to-\infty$. Therefore

(6.32)

$\displaystyle d(H^{*}\theta-\tilde{\theta}_{b_{-}})=0$

where $\tilde{\theta}_{b_{-}}=\theta_{b_{-}}+a$. Since $H^{*}\theta$ and $\tilde{\theta}_{b_{-}}$ are two connections with the same curvature form, and since $H^{1}(U_{-},\mathbb{R})\cong H^{1}(\mathbb{T}^{2},\mathbb{R})\cong\mathbb{R}\oplus\mathbb{R}$, we conclude that

(6.33)

$\displaystyle H^{*}\theta-\tilde{\theta}_{b_{-}}=df+pdx+qdy,$

for some function $f:U_{-}\rightarrow\mathbb{R}$, and constants $p,q\in\mathbb{R}$.

Next, there exists a gauge transformation, that is, a mapping $G:\mathcal{N}_{b_{-}}\rightarrow\mathcal{N}_{b_{-}}$, covering the identity map, given by fiber rotation by $e^{if}$, so that

(6.34)

$\displaystyle G^{*}H^{*}\theta-\tilde{\theta}_{b_{-}}=pdx+qdy.$

Then, by the discussion in Subsection 2.2, there exists a mapping $R:\mathcal{N}_{b_{-}}\rightarrow\mathcal{N}_{b_{-}}$ which is the lift of a rotation on the torus, so that $R^{*}\theta_{b_{-}}=\theta_{b_{-}}-pdx-qdy$. Pulling back (6.34),

(6.35)

$\displaystyle R^{*}G^{*}H^{*}\theta-R^{*}\tilde{\theta}_{b_{-}}=R^{*}(pdx+qdy).$

Since $R$ covers a rotation on the torus, the right hand side is invariant under $R$, so this can be rewritten as

(6.36)

$\displaystyle R^{*}G^{*}H^{*}\theta-\theta_{b_{-}}^{\prime}=0,$

where $\theta_{b_{-}}^{\prime}=\theta_{b_{-}}+O(e^{\delta z})$ as $z\rightarrow-\infty$. Then we define $\Phi^{N}_{-}=H\circ G\circ R$. The $z$ coordinate is not affected because $H$ and $G$ both cover the identity map, and $R$ covers a rotation on the torus.

Next, it follows from (6.19) that the leading terms of the hyperkähler triple on the neck agree with the model hyperkähler triple for $z\ll 0$ (note we can allow $V$ to become negative, the triple is still defined). The same method from the proof of Lemma 3.7 then yields (6.22).

∎