For Schwarzschild solution:
$\mathrm{d} s^2=\left(1-\frac{2m}{r}\right)dt^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d} r^2-r^2(\mathrm{d}\theta^2+sin^2\theta \mathrm{d}\phi^2),$
Introduce tortoise coordinate:
$ v=t+r_*, \quad w=t-r_*,$ where $r_*=r+2mln(r-2m)$,
the metric becomes
$\mathrm{d} s^{2}=\left (1-\frac{2m}{r}\right )\mathrm{d} v \mathrm{d} w-r^{2}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right).$
Next, define two new coordinate
$ V= e^{v/4m},\quad W=e^{w/4m},$
and then choosing
$T=\frac{1}{2}(V+W), \quad X=\frac{1}{2}(V-W),$
the following metric is obtained
$\mathrm{d} s^{2}= \frac{16 m^{2}}{r} e^{ (-r/2 m)} \mathrm{d} T^{2}-\frac{16 m^{2}}{r} e^{ (-r/2 m)} \mathrm{d} X^{2} -r^{2}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right).$
If I have a vector $\partial/\partial r$, and I want to get it component in $\partial/\partial V$ and $\partial/\partial X$, what is the corresponding coefficient, can I calculate $\partial V /\partial r$ and $\partial X/\partial r$?
It seems there are two ways to decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial X}$
(1) directly decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial X}$:
$\frac{\partial}{\partial r}=\frac{\partial V}{\partial r}\frac{\partial}{\partial V}+\frac{\partial X}{\partial r}\frac{\partial}{\partial X}$.
(2) decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial W}$, and combine then to form a $\frac{\partial}{\partial X}$:
$\frac{\partial}{\partial r}=\frac{\partial V}{\partial r}\frac{\partial}{\partial V}+\frac{\partial W}{\partial r}\frac{\partial}{\partial W}=\frac{\partial V}{\partial r}\frac{\partial}{\partial V}+\frac{\partial W}{\partial r}(\frac{\partial}{\partial V}-2\frac{\partial}{\partial X})$.
It could be viewed that the coefficient of $\frac{\partial}{\partial V}$ doesn't match, so one of them must be wrong, which is right?