We next define the “attaching maps” which will be used to construct the manifold $\mathcal{M}$. Let

(6.37)

$\displaystyle DZ_{-}=\Phi^{TY}_{-}\Big((T_{-},T_{-}+1)\times\Nil^{3}_{b_{-}}\Big)\subset X_{b_{-}}^{4},$

using the above coordinates on the end of $X_{b_{-}}^{4}$. Define

(6.38)

$\displaystyle\Psi_{-}:DZ_{-}\rightarrow\mathcal{N}_{m_{0}}^{4}$

by $\Psi_{-}(\Phi^{TY}_{-})^{-1}(z_{-},p)=\Phi^{N}_{-}(z_{-}-2T_{-},p)$.

Simlarly, let

(6.39)

$\displaystyle DZ_{+}=\Phi^{TY}_{-}\Big((T_{+},T_{+}+1)\times\Nil^{3}_{b_{+}}\Big)\subset X_{b_{+}}^{4},$

using the above coordinates on the end of $X_{b_{+}}^{4}$. Define

(6.40)

$\displaystyle\Psi_{+}:DZ_{+}\rightarrow\mathcal{N}_{m_{0}}^{4}$

be defined by $\Psi_{+}(\Phi^{TY}_{+})^{-1}(z_{+},p)=\Phi^{N}_{+}(2T_{+}-z_{+},\psi(p))$, where $\psi_{+}:\Nil^{3}_{b_{+}}\rightarrow\Nil^{3}_{-b_{+}}$ is the diffeomorphism given by

(6.41)

$\displaystyle\psi(x,y,t)=(-x,-y,-t).$

We obtain the manifold $\mathcal{M}$ by gluing the pieces together using the attaching maps:

(6.42)

$$\mathcal{M}\equiv X_{b_{-}}^{4}(T_{-}+1)\bigcup_{\Psi_{-}}\mathcal{N}_{m_{0}}^{4}(-T_{-},T_{+})\bigcup_{\Psi_{+}}X_{b_{+}}^{4}(T_{+}+1).$$

The manifold $\mathcal{M}$ carries an orientation compatible with both Tian-Yau pieces, and we will fix this orientation in the following.

Next, we want the potentials to agree up to the constant term in the damage zones after identifying the corresponding regions by the attaching maps. On $DZ_{-}$ we have

(6.43)		$\displaystyle\Psi_{-}^{*}V_{\beta}$	$\displaystyle=\frac{2\pi}{A}b_{-}(z_{-}-2T_{-})+\beta_{-}+\beta$
(6.44)			$\displaystyle=\frac{2\pi}{A}b_{-}z_{-}+\Big(-\frac{2\pi}{A}b_{-}(2T_{-})+\beta_{-}+\beta\Big),$

which we want to equal to the leading terms of $V_{-}$, so we must have

(6.45)

$\displaystyle 0=-\frac{2\pi}{A}b_{-}(2T_{-})+\beta_{-}+\beta.$

Similarly, on the other damage zone $DZ_{+}$ we have

(6.46)		$\displaystyle\Psi_{+}^{*}V_{\beta}$	$\displaystyle=-\frac{2\pi}{A}b_{+}(2T_{+}-z_{+})+\beta_{+}+\beta$
(6.47)			$\displaystyle=\frac{2\pi}{A}b_{+}z_{+}+\Big(-\frac{2\pi}{A}b_{+}(2T_{+})+\beta_{+}+\beta\Big),$

which we want to equal to the leading terms of $V_{+}$, so we must have

(6.48)

$\displaystyle 0=-\frac{2\pi}{A}b_{+}(2T_{+})+\beta_{+}+\beta.$

To summarize: the gluing procedure requires

(6.49)

$\displaystyle\frac{4\pi b_{\pm}}{A}\cdot T_{\pm}$

$\displaystyle=\beta_{\pm}+\beta.$

Immediately, the above constraints give $1$ free parameter $\beta>0$. So $T_{-}>0$ and $T_{+}>0$ are completely determined by $\beta>0$ in the case $b_{-}>0$ and $b_{+}>0$, i.e.,

(6.50)

$\displaystyle T_{\pm}$

$\displaystyle=\frac{A(\beta+\beta_{\pm})}{4\pi b_{\pm}}.$