Consider a Lorentzian manifold $(\zeta,g)$ with metric: $$g=\frac{dudv}{uv}-\frac{dr^2}{r^2}-\frac{dw^2}{w^2}.$$
For $u,v,w,r,>0$.
Suppose we take a Cauchy foliation of $\zeta,$ called $\mathcal F$, but restrict to the metric to: $h=dudv-dr^2-dw^2.$ Now $\mathcal F$ is embedded in Minkowski space, $(\Bbb M^{3,1},h)$ with Dirac coordinates. Suppose we use the induced measure from $h$ by means of the volume form to transform each leaf in $\mathcal F$ in the region $u,v,w,r \in (0,1)$.
This defines a smooth map $\phi:\mathcal F \to \mathcal H.$ Well, $\mathcal F$ lived in $X=(0,1)^4$ so the target space is dependent upon the measure we use, but the target space will be a $4$-manifold and essentially a transformed hypercube, call it $\mathrm{Image}(X).$
How do you prove the claim that $\phi$ is a smooth map and how do you construct an explicit isometry between $X$ and $\mathrm{Image}(X)$ using $\phi?$
Consider a submanifold $M^2 \subset \zeta$ with metric $g_2=\frac{dudv}{uv}.$ Then we define a Cauchy foliation on $M^2$ as such:
$$ \varphi_s(u)=\mathcal{\exp\bigg(\frac{s}{\log u}\bigg)}. $$
for $s \in \Bbb R$ and $u \in (0,1).$
Then we restrict the metric to $h_2=dudv.$ Now $\varphi_s(u)$ is embedded in $(\Bbb M^{1,1},h_2)$ with Dirac coordinates.
Suppose we use the induced measure from $h$ by means of the volume form to transform each leaf in $\varphi_s(u)$ in the region $u,v \in (0,1):$
$$\Phi(\varphi)(s,r)=\int_{\Omega=(0,1)} u^{r}~\varphi_s(u)~{\frac{du}{u}}=2 \sqrt{\frac{r}{s}}K_1(2\sqrt{rs})$$
for modified Bessel function $K_1.$
I've shown that the operator $\Phi$ defines a linear mapping, a bijection, and is continuous. I can neither show that $\Phi$ is a smooth mapping, nor an isometry. That is what I need help with.
I will prove that the mapping $T$ is linear, continuous and smooth. I will not explicitly construct an isometry at this time.
We are given a spacetime: $$(\zeta,g)$$ which we recognize as Minkowski space in different coordinates. I will change the notation here slightly but the core idea remains intact.
We start by observing that the product space of the foliations:
$$\Omega_{t,r,\theta}(x,y,z)= \varphi_t(x)\varphi_r(y)\varphi_{\theta}(z) \space\space\space\space\space{x,y,z\in(0,1)} \space\space\space{t,r,\theta >0}.$$
Noticeably yields a product space $\Bbb R^{3,3}$ where $(t,r,\theta)$ are time variables and $(x,y,z)$ are space variables. This $6$-manifold is not entirely helpful, so we use a reduction scheme down to $(3+1)$ dimensions by letting $s=t=r=\theta.$ This collapses the time variables down to just one dimension, which is what we want.
Now we are left with a proper Cauchy foliation, $\Omega$, of a $(3+1)$ spacetime, in fact this spacetime is precisely the $\zeta$ we were initially given:
$$ \Omega_s(x,y,z)=\varphi_s(x)\varphi_s(y)\varphi_s(z)\space\space\space\space{s>0}$$
And now we restrict the metric to $h=dudv-dr^2-dw^2$ and hit each leaf with the transform:
$$ \int_{(0,1)^3} \Omega_s(x,z,y)~dxdydz=\bigg(\int_{(0,1)} \Omega_s(x)~dx\bigg)^3= \psi^3(s)$$
We have an explicit function to describe the leaves of the Cauchy foliation and we now have an explicit function for the hyper-volume. This transform maps from the spatial domain to the time domain. Whether that has any physical interpretation, or lack thereof, is irrelevant for this discussion. We can generalize the integral transform to:
$$\int_{(0,1)^3} x^{a}y^{b}z^{c}\Omega_s(x,z,y)~\frac{dxdydz}{xyz}=\chi_s(a,b,c).$$
And so we have a mapping given by the integral transform:
$$T: \Omega \to \chi $$
We have that $T$ is linear because it is a Mellin transform on bounded domain, which we know to be linear. And as $s$ varies continuously on $\Omega$ the corresponding change of $s$ is also continuous on $\chi$. The map is smooth as well, because a smooth change in $s$ on $\Omega$ corresponds to a smooth change of $s$ on $\chi$. This completes the proof that $T$ is smooth.
$\square$