Why find the curl in this paper? I was crossing over this interesting paper http://arxiv.org/abs/hep-th/9404006 when I couldn't realize what happened in the Second Section p. 3: 
They introduced the metric:
$$ds^2 =(V\bar{V})(dt+\overrightarrow{w}d\overrightarrow{x})^2 -(V\bar{V})^{-1}(d\overrightarrow{x})^2 $$ 
 and then tried to find the curl of $\overrightarrow{w}$:$$\overrightarrow{\nabla}\times \overrightarrow{w}=(V\bar{V})^{-1}\overrightarrow{\nabla} \log\frac{V}{\bar{V}}$$ obviously that is intended to help them find $$F_{0i} = E_i =1/2\partial_{\hat{i}}(V+\bar{V})$$
I didn't understand why was this step taken {step of the curl of $w$ or what appears to look like a curl (my second written equation) }and if this step is necessary to be employed later how do we find the curl of it in this case? 
Note that $V$ is a complex function and $w$, they didn't say pretty much what it is.
EDIT: $w$ turned out to be one-form in their previous paper.
 A: This is IWP metric and actually the charged version of Taub-NUT metric with some redefinition of variables. Charged Taub-Nut metric is
\begin{equation}
ds^2 = f (r)(dt + 2N \cos \theta d\phi)^2 − f^{-1}(r)dr^2 − 
(r^2 + N^2)d{\Omega_{2}}^2
\end{equation}
where
\begin{equation}
f(r)= \frac{(r-r_+)(r-r_-)}{r^2+N^2}
\end{equation}
and
\begin{equation}
r_{\pm}=M^2\pm \sqrt{M^2+N^2-4q^2}\,.
\end{equation}
In this equation $q$ is electromagnetic charge and naturally is related to a vector potential. But in general we use Dirac monopole equation for electromagnetic charges, i.e., $\vec\nabla \times\vec A=-\frac{p}{4\pi} \vec\nabla (1/r)$ not the usual $\nabla \cdot \vec B=0$ (since we want to talk about the quantization). Now look at the Taub-NUT metric in the extremal limit where two roots of $f(r)$ are the same. In this limit and by shifting $r$ to $r+M$, you can find that $f^{-1}(r)$ transforms to
\begin{equation}
1+\frac{M^2+2rM+N^2}{r^2}\,.
\end{equation}
Now call this $V\bar V$ where
\begin{equation}
V=1+\frac{M+iN}{r}\,.
\end{equation}
On the hand, if you take $g_{t\phi}$ and consider a 3-vector as
$(0,0,A_{\phi}=g_{t\phi}=2N\cos \theta)$, you can see that this vector satisfies in monopole equation in spherical polar coordinates (you denote this vector by $\omega$).
In short, the curl that you mentioned above won't be completely driven from this metric, but we should consider the monopole equation as well.
