The relationship between the parameters $t$ and $T$ can be determined by studying the matching between the Tian-Yau ends and the neck.

In our setting we need to first normalize the Tian-Yau metrics $\omega_{TY,i}$ on $Z_{i}$ (as defined in Section 7.2). We define

(7.58)

$$\tilde{\omega}_{TY,j}=2^{\frac{-1}{n}}n^{\frac{1}{n}}d_{3-j}^{-\frac{n-1}{n}}\omega_{TY,j}.$$

Then we have

(7.59)

$$\tilde{\omega}_{TY,j}=\frac{(\sqrt{-1})^{n^{2}}}{2^{n}}\Gamma_{0,j}\wedge\bar{\Gamma}_{0,j}.$$

By definition we can write

(7.60)

$$\tilde{\omega}_{TY,j}\equiv dd^{c}\phi_{j}=2\sqrt{-1}\partial\bar{\partial}\phi_{j},$$

where

(7.61)

$$\phi_{j}=\eta_{j}+\psi_{j},$$

with

(7.62)

$$\eta_{j}=\frac{1}{n+1}\cdot k_{3-j}^{\frac{1-n}{n}}n^{\frac{n+1}{n}}(-\log|f_{3-j}|)^{\frac{n+1}{n}},$$

and

(7.63)

$$|\nabla^{k}\psi_{1}|=O(e^{-\delta_{0}(-\log|f_{3-j}|^{2})^{1/2}}),$$

for all $k\geq 0$, where the derivatives and norms are taken with respect to the Tian-Yau metric itself (which is equivalent to taking with respect to the metric $\omega_{Z_{j}}$).

Now on the neck $\mathcal{M}_{T}$ we have the asymptotics of the Kähler potential given in Section 4.2. By the discussion there we identify $\mathcal{M}_{T}$ with an open set in $\mathcal{N}^{0}$, and we can write

(7.64)

$$T^{\frac{n-2}{n}}\omega_{T}=dd^{c}\phi_{T},$$

with

(7.65)

$$\phi_{T}=\begin{cases}\phi_{-}\equiv\varphi_{-}+\psi_{-},\ \ \ \ z<0;\\ \phi_{+}\equiv\varphi_{+}+\psi_{+},\ \ \ \ z>0,\end{cases}$$

where

(7.66)

$$\begin{cases}\varphi_{-}=\frac{1}{n+1}n^{\frac{n+1}{n}}k_{-}^{-\frac{n-1}{n}}(A_{-}-\log|s_{1}/s_{3}|);\\ \varphi_{+}=\frac{1}{n+1}n^{\frac{n+1}{n}}(-k_{+})^{-\frac{n-1}{n}}(A_{+}-\log|s_{2}/s_{3}|),\end{cases}$$

and for $|z|\geq 1$ we have

(7.67)

$$|\psi_{\pm}|=\epsilon(z)+\epsilon_{T}.$$

Now on $\mathcal{M}_{T}$ for $|t|$ small

(7.68)

$$t^{d_{1}}s_{1}=s_{3}f_{2}(x)$$

which gives

(7.69)

$$-d_{1}\log|t|-\log\frac{|s_{1}|}{|s_{3}|}=\log\frac{|s_{3}|}{|s_{1}|}=-\log|f_{2}|.$$

So if we want to graft the metrics on the three components of $\widehat{X}_{0}$ to nearby $\widehat{X}_{t}$, then we need

(7.70)

$$d_{1}\log|t|=-A_{-}.$$

Similarly at the positive end we need

(7.71)

$$d_{2}\log|t|=-A_{+}.$$

This suggests that we should choose

(7.72)

$$|t|=e^{-\frac{1}{d_{1}}A_{-}}=e^{-\frac{1}{d_{2}}A_{+}}.$$

Given $|t|$ small we can find $T$ big so that (7.72) holds. It is not necessary that $T$ is uniquely determined by $t$, but we shall always fix a particular choice for each $t$ throughout this section so that (7.72) holds. With this choice it is easy to see that

(7.73)

$$C^{-1}e^{-\frac{1}{d_{1}d_{2}n}T^{2}}\leq|t|\leq Ce^{-\frac{1}{d_{1}d_{2}n}T^{2}}.$$