Notice the argument above does not essentially require the compactness of $D$, except to solve the equation (4.150) on one slice. Using similar idea can get the expression of the Taub-NUT metric on $\mathbb{C}^{2}$ in terms of Kähler potentials, as mentioned in Section 2.3. Here we take $D$ to be $\mathbb{C}$ with the standard flat structure, and

(4.154)

$$\tilde{\omega}(z)=\frac{\sqrt{-1}}{2}Vdy\wedge d\bar{y};\ \ \ \ h=V,$$

with

(4.155)

$$V=\frac{1}{2r}+T.$$

Suppose we want to find $\phi$ with

(4.156)

$$\omega=dd^{c}\phi,$$

then we first have

(4.157)

$$\phi(z)-\phi(0)=\int_{0}^{z}(\frac{1}{2r}+T)du=\frac{1}{2}r-\frac{1}{2}|y|+\frac{T}{2}z^{2}$$

The equation (4.150) for $z=0$ becomes

(4.158)

$$4\partial_{y}\partial_{\bar{y}}\phi(0)=\tilde{\omega}(0)=\frac{1}{2|y|}+T$$

and a solution is given by

(4.159)

$$\phi(0)=\frac{1}{2}|y|+\frac{T}{4}|y|^{2}$$

So we get

(4.160)

$$\phi=\frac{1}{2}r+\frac{T}{2}z^{2}+\frac{T}{4}|y|^{2}.$$

In terms of the $u_{1},u_{2}$ coordinates we get

(4.161)

$$\phi=\frac{1}{4}(|u_{1}|^{2}+|u_{2}|^{2})+\frac{T}{8}(|u_{1}|^{4}+|u_{2}|^{4}).$$

This agrees with formula (7.61) up to a constant $2$, again caused by the fact that $dd^{c}=2\sqrt{-1}\partial\bar{\partial}$.